# Solved Examples on the Overview, Domain, and Range of Functions

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Unless specified otherwise:
For each of these relations:
(1.) Determine whether the relation is a function or not.
(2.) If it is a function, specify the kind or type of function.
(3.) Write the domain of the relation in set notation.
(4.) Write the domain of the relation in interval notation.
(5.) Write the range of the relation in set notation.
(6.) Write the range of the relation in interval notation.

NOTE: We are concerned with only real numbers.
Let: $D$ = domain; $R$ = range

From Questions (6.) through the rest of the questions in this section; ask this question:
Domain: What are the values of the input, $x$ that will not give any output, $y$?
Remove those values.
The set of all the input values that will give results when those input values are substituted in the functions is the Domain

Notable Notes:
(a.) For rational functions: the denominator must be non-zero.
No division by zero.

$f(x) = \dfrac{numerator}{denominator} \\[5ex] denominator \ne 0 \\[3ex]$ (You cannot divide something by nothing, neither can you divide nothing by nothing.)
So, we have to find all the values of x for which the denominator is not zero.

(b.) For even radicals (even-numbered roots) such as square root, fourth root, sixth root, eighth root, tenth root, etc.; the radicand must be non-negative (zero or positive)
This is because the square root and the even-numbered roots of negative radicals are imaginary numbers.

$f(x) = \sqrt[even-numbered\;root]{radicand} \\[3ex] radicand \ge 0 \\[3ex]$ So, we have to find all the values of x for which the radicand is non-negative.

Range: What are all the likely results that could be got from all those input values?
The set of all those output values that could be got when those input values are substituted in the functions is the Range

Tasks/Questions: Use direct questionining technique if several students are not participating actively in the class.
(1.) What is the difference between: $D(-3, 5)$ and $D = (-3, 5)$
Hint: Which one is: a point? written in interval notation?

(2.) Give students some input values for each relation and ask them to calculate the output values
What will be the output values for those input values?

(3.) Give students some output values and ask them to calculate the input values.
What input values will give those output values?

Please visit: Domain and Range from Graphs (https://openstax.org/books/college-algebra/pages/3-2-domain-and-range)
Scroll down till you see: Finding Domain and Range from Graphs

(1.) {(3, 2), (5, 1), (7, 8)}

It is a function.
It is a bijective function.

$D = \{3, 5, 7\} \\[3ex] R = \{2, 1, 8\}$
(2.) {(7, −3), (−5, 4), (−3, −4), (7, 2)}

It is not a function.
The input, 7 has two different outputs, −3 and 2.
A student cannot make two different grades on the same test.

$D = \{7, -5, -3\} \\[3ex] R = \{-3, 4, -4, 2\}$
(3.) {(4, 3), (3, 3), (−3, 7)}

It is a function.
Two inputs, 4 and 3 can have the same output, 3
It is an onto function.
Two students can make the same grade on a test.

$D = \{4, 3, -3\} \\[3ex] R = \{3, 7\}$
(4.) {(Sam, 3), (Dom, 7), (Math, 12), ( , 10)}

It is a function.
It is a one-to-one function.

$D = \{Sam, Dom, Math\} \\[3ex] R = \{3, 7, 12\}$
(5.) {(Democrat, 0), (Republican, 0), (Independent, 0)}

It is a function.
Two or more inputs, $Democrat$, $Republican$, and $Independent$ can have the same output, $0$
The opposite of what we teach our students in schools - zero collaboration among the lawmakers.
It is an onto function.

$D = \{Democrat, Republican, Independent\} \\[3ex] R = \{0\}$
(6.) $y = x$

It is a function (linear function).
Each input value has an exactly one output value.
It is a one-to-one function.

$D = \{x | x \in \mathbb{R}\} \\[3ex] D = (-\infty, \infty) \\[3ex] R = \{y | y \in \mathbb{R}\} \\[3ex] R = (-\infty, \infty)$
(7.) $y = -x$

It is a function (linear function).
Each input has an exactly one output.
It is a one-to-one function.

$D = \{x | x \:is\: \:a\: \:real\: number\} \\[3ex] D = (-\infty, \infty) \\[3ex] R = \{y | y \in \mathbb{R}\} \\[3ex] R = (-\infty, \infty)$
(8.) $y = x + 4$

It is a function (linear function).
Each input has an exactly one output.
It is a one-to-one function.

$D = \{x | x \in \mathbb{R}\} \\[3ex] D = (-\infty, \infty) \\[3ex] R = \{y | y \:is\: \:a\: \:real\: number\} \\[3ex] R = (-\infty, \infty)$
(9.) $y = 2x - 5$

It is a function (linear function).
Each input has an exactly one output.
It is a one-to-one function.

$D = \{x | x \in \mathbb{R}\} \\[3ex] D = (-\infty, \infty) \\[3ex] R = \{y | y \in \mathbb{R}\} \\[3ex] R = (-\infty, \infty)$
(10.) $y = x^2$

It is a function (quadratic function).
Each input has a unique output.
$D$ = {$x | x \in \mathbb{R}$}. Any input will give an output.
$D = (-\infty, \infty)$
$R$ = {$y | y \ge 0$}. You cannot have any negative result when you square any number.
The square of any number (positive, zero, or negative) will always result in a nonnegative (means: zero and positive) result.
The square of any negative number is a positive result.
The square of zero is zero. Hence, zero is included.
The square of any other input gives a positive output.
$R = [0, \infty)$
(11.) $y = -x^2$

$y = -x^2$ means $y = -1 * x^2$
It is a function (quadratic function).
For each input, there is a unique output.
$D$ = {$x | x \in \mathbb{R}$}. Any input will give an output.
$D = (-\infty, \infty)$
$R$ = {$y | y \le 0$}. We can only have negative and zero output values (nonpositive results).
The square of any number (positive, zero, or negative) will always result in a nonnegative (means zero and positive) result.
Multiply that result by negative $-1$
The square of zero is zero. $0 * -1 = 0$. Zero is included.
The square of any other input gives a positive output. Multiply the result by $-1$. It gives a negative output.
You count from the negative values up to zero, not the other way round.
$R = (-\infty, 0]$
(12.) $y = x^2 + 3$

It is a function (quadratic function).
For each input, there is a unique output.
$D$ = {$x | x \in \mathbb{R}$}. Any input will give an output.
$D = (-\infty, \infty)$
$R$ = {$y | y \ge 3$}.
The square of any number will give a nonnegative result (zero and positive result).
Then, add the result to $3$
The square of zero is zero. $0 + 3$ = $3$
The square of any other input gives a positive output. Add that result to $3$
$3$ is the minimum output that you can ever get.
$R = [3, \infty)$
(13.) $y = x^2 - 5$

It is a function (quadratic function).
For each input, there is a unique output.
$D$ = {$x | x \in \mathbb{R}$}. Any input will give an output.
$D = (-\infty, \infty)$
$R$ = {$y | y \ge -5$}.
The square of any number will give a nonnegative result (zero and positive result).
Then, add the result to $-5$
The square of zero is zero. $0 + (-5)$ = $-5$
The square of any other input gives a positive output. Add that result to $-5$
$3$ is the minimum output that you can ever get.
$R = [-5, \infty)$
(14.) $y = x^3$
The same answer applies to $y = x$, $y = x^5$, $y = x^7$, ...
Ask students to explain these functions with odd exponents have the same domain and range.

It is a function (cubic function).

$D = \{x | x \in \mathbb{R}\} \\[3ex] D = (-\infty, \infty) \\[3ex] R = \{y | y \in \mathbb{R}\} \\[3ex] R = (-\infty, \infty)$
(15.) $y = x^4$
The same answer applies to $y = x^2$, $y = x^6$, $y = x^8$, ...
Ask students to explain these functions with even exponents have the same domain and range.

It is a function (quartic function).

$D = \{x | x \in \mathbb{R}\} \\[3ex] D = (-\infty, \infty) \\[3ex] R = \{y | y \ge 0\} \\[3ex] R = [0, \infty)$
(16.) $y = \dfrac{x + 3}{5}$

It is a function (linear function).

$y = \dfrac{x + 3}{5} = \dfrac{x}{5} + \dfrac{3}{5} \\[5ex] D = \{x | x \in \mathbb{R}\} \\[3ex] D = (-\infty, \infty) \\[3ex] R = \{y | y \:is\: real\} \\[3ex] R = (-\infty, \infty)$
(17.) $y = \dfrac{2x - 7}{3}$

It is a function (linear function).

$y = \dfrac{2x - 7}{3} = \dfrac{2x}{3} - \dfrac{7}{3} \\[5ex] D = \{x | x \in \mathbb{R}\} \\[3ex] D = (-\infty, \infty) \\[3ex] R = \{y | y \in \mathbb{R}\} \\[3ex] R = (-\infty, \infty)$
(18.) $y = \dfrac{3}{x}$

It is a function (rational function).
The denominator of rational functions must be non-zero.
We cannot divide anything by nothing.
Dividing anything (something or nothing) by nothing does not give us a real number.
It is undefined.
We are dealing with only real numbers.

$D = \{x | x \in \mathbb{R}; x \ne 0\} \\[3ex] D = (-\infty, 0) \bigcup (0, \infty) \\[3ex]$ To find the range, let us assume the output is $0$

$y = \dfrac{3}{x} \\[5ex] 0 = \dfrac{3}{x} \\[5ex] \dfrac{3}{x} = 0 \\[5ex] 3 = x * 0 \\[3ex] 3 = 0? \:This\: is\: a\: contradiction \\[3ex]$ $0$ can never be an output.
Any other number can be an output.

$R = \{y | y \in \mathbb{R}; y \ne 0\} \\[3ex] R = (-\infty, 0) \bigcup (0, \infty)$
(19.) $xy = -7$

$xy = -7 \\[3ex] y = -\dfrac{7}{x} \\[5ex]$ It is a function (rational function).
The denominator of rational functions must be non-zero.
We cannot divide anything by nothing.
Dividing anything (something or nothing) by nothing does not give us a real number.
It is undefined.
We are dealing with only real numbers.

$D = \{x | x \in \mathbb{R}; x \ne 0\} \\[3ex] D = (-\infty, 0) \bigcup (0, \infty) \\[3ex]$ To find the range, let us assume the output is $0$

$y = -\dfrac{7}{x} \\[5ex] 0 = -\dfrac{7}{x} \\[5ex] -\dfrac{7}{x} = 0 \\[5ex] -7 = x * 0 \\[3ex] -7 = 0? \:This\: is\: a\: contradiction \\[3ex]$ $0$ can never be an output.
Any other number can be an output.

$R = \{y | y \in \mathbb{R}; y \ne 0\} \\[3ex] R = (-\infty, 0) \bigcup (0, \infty)$
(20.) $y = \dfrac{5}{x - 4}$

It is a function (rational function).
The denominator of rational functions must be non-zero.

$x - 4 \ne 0 \\[3ex] x \ne 4 \\[3ex] D = \{x | x \in \mathbb{R}; x \ne 4\} \\[3ex] D = (-\infty, 4) \bigcup (4, \infty) \\[3ex]$ To find the range, let us assume the output is $0$

$y = \dfrac{5}{x - 4} \\[5ex] 0 = \dfrac{5}{x - 4} \\[5ex] \dfrac{5}{x - 4} = 0 \\[5ex] 5 = (x - 4) * 0 \\[3ex] 5 = 0? \:This\: is\: a\: contradiction \\[3ex]$ $0$ can never be an output.
Any other number can be an output.

$R = \{y | y \in \mathbb{R}; y \ne 0\} \\[3ex] R = (-\infty, 0) \bigcup (0, \infty)$

(21.) $y = -\dfrac{4}{2x + 5}$

It is a function (rational function).
The denominator of rational functions must be non-zero.

$2x + 5 \ne 0 \\[3ex] 2x \ne -5 \\[3ex] x \ne -\dfrac{5}{2} \\[5ex] D = \{x | x \in \mathbb{R}; x \ne -\dfrac{5}{2}\} \\[5ex] D = \left(-\infty, -\dfrac{5}{2}\right) \bigcup \left(-\dfrac{5}{2}, \infty\right) \\[5ex]$ To find the range, let us assume the output is $0$

$y = -\dfrac{4}{2x + 5} \\[5ex] 0 = -\dfrac{4}{2x + 5} \\[5ex] -\dfrac{4}{2x + 5} = 0 \\[5ex] -4 = (2x + 5) * 0 \\[3ex] -4 = 0? \:This\: is\: a\: contradiction \\[3ex]$ $0$ can never be an output.
Any other number can be an output.

$R = \{y | y \in \mathbb{R}; y \ne 0\} \\[3ex] R = (-\infty, 0) \bigcup (0, \infty)$
(22.) $y = \pm \sqrt{x}$

It is not a function because an input value can have at least two output values.
Examples: $\{(1, -1), (1, 1), (4, -2), (4, 2), ...\}$

To find the domain, the radicand (the value inside the square root radical) must not be negative.
Square root of negative numbers are not real numbers. They are imaginary numbers.
We are only dealing with real numbers.
So, the radicand must be a nonnegative number (zero and positive number).
That means that the radicand must be at least zero.

$x \ge 0 \\[3ex] D = \{x : x \ge 0\} \\[3ex] D = [0, \infty) \\[3ex]$ For the range, due to the plus-minus sign in front of the radical; the range includes all real numbers.

$R = \{y | y \in \mathbb{R}\} \\[3ex] R = (-\infty, \infty)$
(23.) $y = \sqrt{x}$
The same answer applies to $y = \sqrt[4]{x}$, $y = \sqrt[6]{x}$, $y = \sqrt[10]{x}$, ...
Ask students to explain why the functions with even-th roots (even-numbered roots) have the same domain and range.

It is a function (positive square root function). It is also a one-to-one function.
Each input has exactly one output.

To find the domain, the radicand (the value inside the square root radical) must not be negative.
Square root of negative numbers are not real numbers. They are imaginary numbers.
We are only dealing with real numbers.
So, the radicand must be a nonnegative number (zero and positive number).
That means that the radicand must be at least zero.

$x \ge 0 \\[3ex] D = \{x : x \ge 0\} \\[3ex] D = [0, \infty) \\[3ex]$ For the range, there is only the plus sign in front of the radical.
The smallest "real number" output value occurs for the smallest input value.
$\sqrt{0} = 0$
The smallest output value is $0$
The range can only include nonnegative numbers (zero and positive numbers).

$R = \{y | y \ge 0\} \\[3ex] R = [0, \infty)$
(24.) $y = \sqrt{x - 9}$

It is a function (positive square root function). It is also a one-to-one function.
Each input has exactly one output.

To find the domain, the radicand (the value inside the square root radical) must not be negative.
Square root of negative numbers are not real numbers. They are imaginary numbers.
We are only dealing with real numbers.
So, the radicand must be a nonnegative number (zero and positive number).
That means that the radicand must be at least zero.

$x + 9 \ge 0 \\[3ex] x \ge 0 + 9 \\[3ex] x \ge 9 \\[3ex] D = \{x : x \ge 9\} \\[3ex] D = [9, \infty) \\[3ex]$ For the range, there is only the plus sign in front of the radical.
The smallest "real number" output value occurs for the smallest input value.
The smallest input value is $9$

$x - 9 \\[3ex] 9 - 9 = 0 \\[3ex] \sqrt{0} = 0 \\[3ex]$ The smallest output value is $0$
The range can only include nonnegative numbers (zero and positive numbers).

$R = \{y | y \ge 0\} \\[3ex] R = [0, \infty)$
(25.) $y = \sqrt{2x + 12}$

It is a function (positive square root function). It is also a one-to-one function.
Each input has exactly one output.

To find the domain, the radicand (the value inside the square root radical) must not be negative.
Square root of negative numbers are not real numbers. They are imaginary numbers.
We are only dealing with real numbers.
So, the radicand must be a nonnegative number (zero and positive number).
That means that the radicand must be at least zero.

$2x + 12 \ge 0 \\[3ex] 2x \ge 0 -12 \\[3ex] 2x \ge -12 \\[3ex] x \ge -\dfrac{12}{2} \\[5ex] x \ge -6 \\[3ex] D = \{x : x \ge -6\} \\[3ex] D = [-6, \infty) \\[3ex]$ For the range, there is only the plus sign in front of the radical.
The smallest "real number" output value occurs for the smallest input value.
The smallest input value is $-6$

$2 * -6 + 12 \\[3ex] -12 + 12 = 0 \\[3ex] \sqrt{0} = 0 \\[3ex]$ The smallest output value is $0$
The range can only include nonnegative numbers (zero and positive numbers).

$R = \{y | y \ge 0\} \\[3ex] R = [0, \infty)$
(26.) $y = \sqrt{x + 7}$

It is a function (positive square root function). It is also a one-to-one function.
Each input has exactly one output.

To find the domain, the radicand (the value inside the square root radical) must not be negative.
Square root of negative numbers are not real numbers. They are imaginary numbers.
We are only dealing with real numbers.
So, the radicand must be a nonnegative number (zero and positive number).
That means that the radicand must be at least zero.

$x + 7 \ge 0 \\[3ex] x \ge 0 - 7 \\[3ex] x \ge -7 \\[3ex] D = \{x : x \ge -7\} \\[3ex] D = [-7, \infty) \\[3ex]$ For the range, there is only the plus sign in front of the radical.
The smallest "real number" output value occurs for the smallest input value.
The smallest input value is $-7$

$x + 7 \\[3ex] -7 + 7 = 0 \\[3ex] \sqrt{0} = 0 \\[3ex]$ The smallest output value is $0$
The range can only include nonnegative numbers (zero and positive numbers).

$R = \{y | y \ge 0\} \\[3ex] R = [0, \infty)$
(27.) $y = \sqrt{x + 7} - 3$

It is a function (positive square root function). It is also a one-to-one function.
Each input has exactly one output.

To find the domain, the radicand (the value inside the square root radical) must not be negative.
Square root of negative numbers are not real numbers. They are imaginary numbers.
We are only dealing with real numbers.
So, the radicand must be a nonnegative number (zero and positive number).
That means that the radicand must be at least zero.

$x + 7 \ge 0 \\[3ex] x \ge 0 - 7 \\[3ex] x \ge -7 \\[3ex] D = \{x : x \ge -7\} \\[3ex] D = [-7, \infty) \\[3ex]$ For the range, there is only the plus sign in front of the radical.
The smallest input value for the radicand is $9$

$x + 7 \\[3ex] -7 + 7 = 0 \\[3ex] \sqrt{0} = 0 \\[3ex]$ To get the smallest "real number" output value, we have to do $0 - 3 = -3$
The $-3$ is the number outside the square root. We have to consider it.
The smallest output value is $-3$

$R = \{y | y \ge -3\} \\[3ex] R = [-3, \infty)$
(28.) $y = |x|$

It is an absolute value function. It is not a one-to-one function.
Two input values can have the same output.

To find the domain, we have to consider the range of numbers that can be inside the absolute value which would give a result.
Any number can be inside that absolute value and still give a result.

$D = \{x | x \in \mathbb{R}\} \\[3ex] D = (-\infty, \infty) \\[3ex]$ The range can only be a nonnegative number (zero and positive numbers only).
The term inside the absolute value (the domain) can be zero, positive, or negative.
However, the absolute value (the result - the range) can only be nonnegative.

$R = \{y | y \ge 0\} \\[3ex] R = [0, \infty)$
(29.) $y = -|x|$

It is an absolute value function. It is not a one-to-one function.
Two input values can have the same output.

To find the domain, we have to consider the range of numbers that can be inside the absolute value which would give a result.
Any number can be inside that absolute value and still give a result.

$D = \{x | x \in \mathbb{R}\} \\[3ex] D = (-\infty, \infty) \\[3ex]$ The range can only be a nonnegative number (zero and positive numbers only).
The term inside the absolute value (the domain) can be zero, positive, or negative.
The absolute value (the result - the range) can only be nonnegative.
But, we have to consider the negative $-1$ outside the absolute value.
This means that any result we have, would be multiplied by $-1$
That makes all our results to be nonpositive (zero and negative numbers only).

$R = \{y | y \le 0\} \\[3ex] R = (-\infty, 0]$
(30.) $y = |x| + 7$

It is an absolute value function. It is not a one-to-one function.
Two input values can have the same output value.

To find the domain, we have to consider the range of numbers that can be inside the absolute value which would give a result, and see if that result added to $7$ would give a result.
Any number can be inside that absolute value and still give a result.
That result can still be added to $7$ and still produce another result.

$D = \{x | x \in \mathbb{R}\} \\[3ex] D = (-\infty, \infty) \\[3ex]$ The term inside the absolute value (the domain) can be zero, positive, or negative.
However, the absolute value (the result) can only be nonnegative.
To find the range, we also have to consider the $7$ outside the absolute value.
To find the minimum result, we have to ask this question - what is the minimum result from that absolute value?
The minimum result from the absolute value is $0$
We now have to add it to $7$
$0 + 7 = 7$
This means that the minimum result is $7$

$R = \{y | y \ge 7\} \\[3ex] R = [7, \infty)$
(31.) $y = -|x| + 7$

It is an absolute value function. It is not a one-to-one function.
Two input values can have the same output value.

To find the domain, we have to consider the range of numbers that can be inside the absolute value which would give a result, multiplied by $-1$ and give a result, and see if that result added to $7$ would give a result.
Any number can be inside that absolute value and still give a result.
That result can be multiplied by $-1$ and give another result.
That result can still be added to $7$ and still produce another result.

$D = \{x | x \in \mathbb{R}\} \\[3ex] D = (-\infty, \infty) \\[3ex]$ The term inside the absolute value (the domain) can be zero, positive, or negative.
The absolute value (the result) can only be nonnegative.
But, we have to consider the negative $-1$ outside the absolute value.
This means that any result we have, would be multiplied by $-1$
That makes all our results to be nonpositive (zero and negative numbers only).
Then, we have to add the result to the $7$ outside the absolute value.
Let us begin by asking this question - what is the minimum result from the absolute value?
The minimum result from the absolute value is $0$

$-1 * 0 = 0 \\[3ex] 0 + 7 = 7 \\[3ex]$ To see the "entire" range of results, we have to test more values.
Let us test for these values: $-1$ and $1$

$Say\:\:x = -1 \\[3ex] |-1| = 1 \\[3ex] -1 * 1 = -1 \\[3ex] -1 + 7 = 6 \\[3ex] Say\:\:x = 1 \\[3ex] |1| = 1 \\[3ex] -1 * 1 = -1 \\[3ex] -1 + 7 = 6 \\[3ex]$ Let us test for two more values: $-2$ and $2$

$Say\:\:x = -2 \\[3ex] |-2| = 2 \\[3ex] -1 * 2 = -2 \\[3ex] -2 + 7 = 5 \\[3ex] Say\:\:x = 2 \\[3ex] |2| = 2 \\[3ex] -1 * 2 = -2 \\[3ex] -2 + 7 = 5 \\[3ex]$ This means that the "maximum" result is $7$

Do you think we shall have any minimum result?

$R = \{y | y \le 7\} \\[3ex] R = (-\infty, 7]$
(32.) $y = |x| - 7$

It is an absolute value function. It is not a one-to-one function.
Two input values can have the same output value.

To find the domain, we have to consider the range of numbers that can be inside the absolute value which would give a result, and see if $7$ subtracted from that result, gives a result.
Any number can be inside that absolute value and still give a result.
That result can still be added to $7$ and still produce another result.

$D = \{x | x \in \mathbb{R}\} \\[3ex] D = (-\infty, \infty) \\[3ex]$ The term inside the absolute value (the domain) can be zero, positive, or negative.
However, the absolute value (the result) can only be nonnegative.
To find the range, we also have to consider the $7$ outside the absolute value.
To find the minimum result, we have to ask this question - what is the minimum result from that absolute value?
The minimum result from the absolute value is $0$
We now have to subtract $7$ from $0$
$0 - 7 = -7$
This means that the minimum result is $-7$

$R = \{y | y \ge -7\} \\[3ex] R = [-7, \infty)$
(33.) $y = -|x| - 7$

It is an absolute value function. It is not a one-to-one function.
Two input values can have the same output value.

To find the domain, we have to consider the range of numbers that can be inside the absolute value which would give a result, multiplied by $-1$ and give a result, and see if $7$ subtracted from that result, would give a result.
Any number can be inside that absolute value and still give a result.
That result can be multiplied by $-1$ and give another result.
$7$ can be subtracted from that result, and still produce another result.

$D = \{x | x \in \mathbb{R}\} \\[3ex] D = (-\infty, \infty) \\[3ex]$ The term inside the absolute value (the domain) can be zero, positive, or negative.
The absolute value (the result) can only be nonnegative.
But, we have to consider the negative $-1$ outside the absolute value.
This means that any result we have, would be multiplied by $-1$
That makes all our results to be nonpositive (zero and negative numbers only).
Then, we have to add the result to the $7$ outside the absolute value.
Let us begin by asking this question - what is the minimum result from the absolute value?
The minimum result from the absolute value is $0$

$-1 * 0 = 0 \\[3ex] 0 - 7 = -7 \\[3ex]$ To see the "entire" range of results, we have to test more values.
Let us test for these values: $-1$ and $1$

$Say\:\:x = -1 \\[3ex] |-1| = 1 \\[3ex] -1 * 1 = -1 \\[3ex] -1 - 7 = -8 \\[3ex] Say\:\:x = 1 \\[3ex] |1| = 1 \\[3ex] -1 * 1 = -1 \\[3ex] -1 - 7 = -8 \\[3ex]$ Let us test for two more values: $-2$ and $2$

$Say\:\:x = -2 \\[3ex] |-2| = 2 \\[3ex] -1 * 2 = -2 \\[3ex] -2 - 7 = -9 \\[3ex] Say\:\:x = -2 \\[3ex] |-2| = 2 \\[3ex] -1 * 2 = -2 \\[3ex] -2 - 7 = -9 \\[3ex]$ This means that the "maximum" result is $-7$

Do you think we shall have any minimum result?

$R = \{y | y \le -7\} \\[3ex] R = (-\infty, -7]$
(34.) $x = y^2 - 3$

This implies that:

$x + 3 = y^2 \\[3ex] y^2 = x + 3 \\[3ex] y = \pm \sqrt{x + 3} \\[3ex]$ It is not a function because an input value can have at least two output values.
Examples: $\{(1, -2), (1, 2), (6, -3), (6, 3), ...\}$

To find the domain, the radicand (the value inside the square root radical) must not be negative.
Square root of negative numbers are not real numbers. They are imaginary numbers.
We are only dealing with real numbers.
So, the radicand must be a nonnegative number (zero and positive number).
That means that the radicand must be at least zero.

$x + 3 \ge 0 \\[3ex] x \ge 0 - 3 \\[3ex] x \ge -3 \\[3ex] D = \{x : x \ge -3\} \\[3ex] D = [-3, \infty) \\[3ex]$ For the range, due to the plus-minus sign in front of the radical; the range includes all real numbers.

$R = \{y | y \in \mathbb{R}\} \\[3ex] R = (-\infty, \infty)$
(35.) $f(x) = \dfrac{1}{x^2 - x - 6}$

It is a rational function.
First, let us deal with the denominator.
The denominator should not be equal to zero.
The denominator is quadratic. It has two roots.
So, we have to find those two roots.
The denominator should not be equal to those roots
Those roots should not be in our domain because it will give a denominator of zero.
A denominator of zero is an "undefined" function.

$x^2 - x - 6 \ne 0 \\[3ex] (x + 2)(x - 3) \ne 0 \\[3ex] x + 2 \ne 0 \:\:\:OR\:\:\: x - 3 \ne 0 \\[3ex] x \ne -2 \:\:\:OR\:\:\: x \ne 3 \\[3ex]$ Second, let us deal with the numerator.
The numerator is just $1$. It is a constant.
That is not a problem.
So, our domain will include all real numbers besides $-2$ and $3$

$D = \{x | x \ne -2, x \ne 3\} \\[3ex] D = (-\infty, -2) \bigcup (-2, 3) \bigcup (3, \infty) \\[3ex]$ To find the range, we have to be "extra careful" here
First, let us assume our output to be $0$

$y = \dfrac{1}{x^2 - x - 6} \\[5ex] 0 = \dfrac{1}{x^2 - x - 6} \\[5ex] \dfrac{1}{x^2 - x - 6} = 0 \\[5ex] 1 = (x^2 - x - 6) * 0 \\[3ex] 1 = 0? \:This\: is\: a\: contradiction \\[3ex]$ $0$ can never be an output.

Second, let us note that we should only get real numbers as our results.
Consider the fact that the denominator is quadratic.
The discriminant should not be negative because it will lead to complex number results.
So, let us calculate the discriminant.

$x^2 - x - 6 \\[3ex] Compare:\:\:\: ax^2 + bx + c \\[3ex] a = 1, b = -1, c = -6 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = (-1)^2 - 4(1)(-6) \\[3ex] = 1 + 24 \\[3ex] = 25 \\[3ex]$ OR
Better still: let us graph it and determine the range from the graph.

The range will never be zero. It can approach it but can never be it.
The range will never be those y-values in-between the red braces.
It will include everything else.

$-0.16666666666667 = -\dfrac{4}{25} \\[5ex] R = \{y | y \le -\dfrac{4}{25} \:\:\:OR\:\:\: y \gt 0\} \\[3ex] R = \left(-\infty, -\dfrac{4}{25}\right) \bigcup (0, \infty)$
(36.) $f(x) = \sqrt{3 - 5x}$

It is a function (positive square root function).
It is also a one-to-one function.
Each input has an exactly one output.

To find the domain, the radicand (the value inside the square root radical) must not be negative.
Square root of negative numbers are not real numbers. They are imaginary numbers.
We are only dealing with real numbers.
So, the radicand must be a nonnegative number (zero and positive number).
That means that the radicand must be at least zero.

$3 - 5x \ge 0 \\[3ex] 3 - 0 \ge 5x \\[3ex] 3 \ge 5x \\[3ex] 5x \le 3 \\[3ex] x \le \dfrac{3}{5} \\[5ex] D = \{x : x \le \dfrac{3}{5}\} \\[5ex] D = \left(-\infty, \dfrac{3}{5}\right] \\[5ex]$ For the range, there is only the plus sign in front of the radical.
The smallest "real number" output value occurs for the smallest input value.
The smallest input value is $\dfrac{3}{5}$

$3 - 5x \\[3ex] 3 - 5 * \dfrac{3}{5} \\[5ex] 3 - 3 \\[3ex] 0 \\[3ex] \sqrt{0} = 0 \\[3ex]$ The smallest output value is $0$
The range can only include nonnegative numbers (zero and positive numbers).

$R = \{y | y \ge 0\} \\[3ex] R = [0, \infty)$
(37.) $f(x) = \sqrt{3 - 5x} + 3$

It is a function (positive square root function). It is also a one-to-one function.
Each input has an exactly one output.

To find the domain, the radicand (the value inside the square root radical) must not be negative.
Square root of negative numbers are not real numbers. They are imaginary numbers.
We are only dealing with real numbers.
So, the radicand must be a nonnegative number (zero and positive number).
That means that the radicand must be at least zero.

$3 - 5x \ge \\[3ex] 3 - 0 \ge 5x \\[3ex] 3 \ge 5x \\[3ex] 5x \le 3 \\[3ex] x \le \dfrac{3}{5} \\[5ex] D = \{x : x \le \dfrac{3}{5}\} \\[5ex] D = \left(-\infty, \dfrac{3}{5}\right] \\[5ex]$ For the range, there is only the plus sign in front of the radical.
The smallest "real number" output value occurs for the smallest input value.
The smallest input value is $\dfrac{3}{5}$

$3 - 5x \\[3ex] 3 - 5 * \dfrac{3}{5} \\[5ex] 3 - 3 \\[3ex] 0 \\[3ex] \sqrt{0} = 0 \\[3ex]$ To get the smallest output value, we have to do $0 + 3 = 3$
The $3$ is the number outside the square root. We have to consider it.
The smallest output value is $3$

$R = \{y | y \ge 3\} \\[3ex] R = [3, \infty)$
(38.) $f(x) = \dfrac{3}{\sqrt{16 + x}}$

First, let us deal with the denominator.
The denominator should not be equal to zero.
The denominator has a square root.
The radicand should be greater than or equal to zero.
So, we have several conditions here.
And we have to consider those conditions.
Let us work from "inside" to "outside" - "an inside-outside approach".
Condition 1: The radicand should be greater than or equal to zero.

$16 + x \ge 0 \\[3ex] x \ge 0 - 16 \\[3ex] x \ge -16 \\[3ex]$ Condition 2: The denominator should not be equal to zero.

$\sqrt{16 + x} \ne 0 \\[3ex]$ Based on Condition 1, if $x = -16$, that will violate Condition 2.

$If\:\:x = -16 \\[3ex] \sqrt{16 + x} \\[3ex] = \sqrt{16 + -16} \\[3ex] = \sqrt{16 - 16} \\[3ex] = \sqrt{0} \\[3ex] = 0 \\[3ex]$ Based on Condition 2, $\sqrt{16 + x} \ne 0$
So, considering our denominator; only $x \gt -16$ works.

Second, let us deal with the numerator.
The numerator is just $3$. It is a constant.
That is not a problem.
So, our domain will be:

$D = \{x | x \gt -16\} \\[3ex] D = (-16, \infty) \\[3ex]$ To find the range, we have to be "extra careful" here
First, let us assume our output to be $0$

$y = \dfrac{3}{\sqrt{16 + x}} \\[5ex] 0 = \dfrac{3}{\sqrt{16 + x}} \\[5ex] \dfrac{3}{\sqrt{16 + x}} = 0 \\[5ex] 3 = \sqrt{16 + x} * 0 \\[3ex] 3 = 0? \:This\: is\: a\: contradiction \\[3ex]$ $0$ can never be an output.
Second, we cannot have any negative results.
Why?
Assume our result is $-1$

$y = \dfrac{3}{\sqrt{16 + x}} \\[5ex] -1 = \dfrac{3}{\sqrt{16 + x}} \\[5ex] \dfrac{3}{\sqrt{16 + x}} = -1 \\[5ex] \sqrt{16 + x} * -1 = 3 \\[3ex] \sqrt{16 + x} = \dfrac{3}{-1} \\[3ex] \sqrt{16 + x} = -3 \\[3ex]$ Keep in mind that we are not trying to solve for the input, $x$
We are just trying to see what are "allowable" as our results.
So, the question to ask is this:
Can the square root of any real number within our domain give a negative result?
No.
So, the range will also exclude negative results.

$R = \{y | y \gt 0\} \\[3ex] R = (0, \infty)$
(39.) $y = -(x + 7)^2$

$y = -(x + 7)^2$ means $y = -1 * (x + 7)^2$
It is a function (quadratic function).
For each input, there is a unique output.
$D$ = {$x | x \in \mathbb{R}$}. Any input will give an output.
$D = (-\infty, \infty)$
The square of any number will give a nonnegative result (zero and positive result).
This implies that the result of (x + 7)^2 will be nonnegative.
Then, multiply the result by $-1$
This means that our results can only be nonpositive (zero and negative results)
$0$ is the maximum output that we can get.

$R = \{y | y \le 0\} \\[3ex] R = (-\infty, 0]$
(40.) $y = (x + 7)^2$

It is a function (quadratic function).
For each input, there is a unique output.
$D$ = {$x | x \in \mathbb{R}$}. Any input will give an output.
$D = (-\infty, \infty)$
The square of any number will give a nonnegative result (zero and positive result).
This implies that the result of (x + 7)^2 will be nonnegative.
$0$ is the minimum output that we can get.

$R = \{y | y \ge 0\} \\[3ex] R = [0, \infty)$

(41.) $y = -(x + 7)^2 - 12$

$y = -(x + 7)^2 - 12$ means $y = -1 * (x + 7)^2 - 12$
It is a function (quadratic function).
For each input, there is a unique output.
$D$ = {$x | x \in \mathbb{R}$}. Any input will give an output.
$D = (-\infty, \infty)$
The square of any number will give a nonnegative result (zero and positive result).
This implies that the result of (x + 7)^2 will be nonnegative.
Then, multiply the result by $-1$
This means that our results will now be nonpositive (zero and negative results)
$0$ is the "maximum" result so far.
Then, subtract $12$ from the result.
$0 - 12 = -12$
$-12$ is the maximum result we can get from this function.

$R = \{y | y \le -12\} \\[3ex] R = (-\infty, -12]$
(42.) $y = (x - 7)^2 + 12$

It is a function (quadratic function).
For each input, there is a unique output.
$D$ = {$x | x \in \mathbb{R}$}. Any input will give an output.
$D = (-\infty, \infty)$
The square of any number will give a nonnegative result (zero and positive result).
This implies that the result of (x - 7)^2 will be nonnegative.
$0$ is the minimum output that we can get.
Then, add $12$ to the result.
$0 + 12 = 12$
$12$ is the minimum result we can get from this function.

$R = \{y | y \ge 12\} \\[3ex] R = [12, \infty)$
(43.) $y = 9 - \dfrac{7}{x}$

It is a function (rational function).
The denominator of rational functions must be non-zero.
We cannot divide anything by nothing.
Dividing anything (something or nothing) by nothing does not give us a real number.
It is undefined.
We are dealing with only real numbers.

$D = \{x | x \in \mathbb{R}; x \ne 0\} \\[3ex] D = (-\infty, 0) \bigcup (0, \infty) \\[3ex]$ To find the range, let us assume the output is $0$

$y = 9 - \dfrac{7}{x} \\[5ex] 0 = 9 - \dfrac{7}{x} \\[5ex] \dfrac{7}{x} = 9 \\[5ex] 7 = 9x \\[3ex] 9x = 7 \\[3ex] x = \dfrac{7}{9} \\[5ex]$ An input of $\dfrac{7}{9}$ will give an output of $0$.

So, $0$ can be an output.

What do you think we should do?

Let us assume the output is $9$

$y = 9 - \dfrac{7}{x} \\[5ex] 9 = 9 - \dfrac{7}{x} \\[5ex] \dfrac{7}{x} = 9 - 9 \\[5ex] \dfrac{7}{x} = 0 \\[5ex] 7 = x * 0 \\[3ex] 7 = 0? \:\:\:This\:\:\: is\:\:\: a\:\:\: contradiction \\[3ex]$ $9$ can never be an output.
Any other number can be an output.

$R = \{y | y \in \mathbb{R}; y \ne 9\} \\[3ex] R = (-\infty, 9) \bigcup (9, \infty) \\[3ex]$
(44.) $y = |3 - x|$

It is an absolute value function. It is not a one-to-one function.
Two input values can have the same output.

To find the domain, we have to consider the range of numbers that can be inside the absolute value which would give a result.
Any number can be inside that absolute value, subtracted from $3$, and still give a result.

$D = \{x | x \in \mathbb{R}\} \\[3ex] D = (-\infty, \infty) \\[3ex]$ The range can only be a nonnegative number (zero and positive numbers only).
The term inside the absolute value (the domain) can be zero, positive, or negative.
However, the absolute value (the result - the range) can only be nonnegative.

$R = \{y | y \ge 0\} \\[3ex] R = [0, \infty)$
(45.) ACT A function that is defined by the set of ordered pairs {(2, 1), (4, 2), (6, 3)} has domain {2, 4, 6}.
What is the domain of the function defined by the set of ordered pairs {(0, 2), (2, 2), (3, -2)}?

$A.\;\; \{2\} \\[3ex] B.\;\; \{-2, 2\} \\[3ex] C.\;\; \{-2, 0, 3\} \\[3ex] D.\;\; \{0, 2, 3\} \\[3ex] E.\;\; \{-2, 0, 2, 3\} \\[3ex]$

$For\;\;the\;\;set:\;\; \{(0, 2), (2, 2), (3, -2)\} \\[3ex] D = \{0, 2, 3\}$
(46.) ACT Given that the function $f$ defined as $f(x) = 5 - 3x$ has domain $\{-1, 0, 2\}$, what is the range of $f$?

$A.\:\: \{-2, 0, 4\} \\[3ex] B.\:\: \{-1, 2, 8\} \\[3ex] C.\:\: \{-1, 5, 8\} \\[3ex] D.\:\: \{2, 5, 8\} \\[3ex] E.\:\: \{2, 5, 11\} \\[3ex]$

$f(x) = 5 - 3x \\[3ex] D = \{-1, 0, 2\} \\[3ex] f(-1) = 5 - 3(-1) \\[3ex] f(-1) = 5 + 3 \\[3ex] f(-1) = 8 \\[3ex] f(0) = 5 - 3(0) \\[3ex] f(0) = 5 - 0 \\[3ex] f(0) = 5 \\[3ex] f(2) = 5 - 3(2) \\[3ex] f(2) = 5 - 6 \\[3ex] f(2) = -1 \\[3ex] R = \{8, 5, -1\} \\[3ex] R = \{-1, 5, 8\}$
(47.)

$(a.)\;\; f(x) = \sqrt{5 + 2x} \\[3ex] (b.)\;\; f(x) = \sqrt{2x - 5} \\[3ex] (c.)\;\; g(h(x)) \;\; if: \\[3ex] g(x) = \sqrt{3x - 4} \\[3ex] h(x) = 3x + 5 \\[3ex]$

(a.)
It is a function (positive square root function).
It is also a one-to-one function.
Each input has an exactly one output.

To find the domain, the radicand (the value inside the square root radical) must not be negative.
Square root of negative numbers are not real numbers. They are imaginary numbers.
We are only dealing with real numbers.
So, the radicand must be a nonnegative number (zero and positive number).
That means that the radicand must be at least zero.

$5 + 2x \ge 0 \\[3ex] 2x \ge 0 - 5 \\[3ex] 2x \ge -5 \\[3ex] x \ge -\dfrac{5}{2} \\[5ex] D = \{x : x \ge -\dfrac{5}{2}\} \\[5ex] D = \left[-\dfrac{5}{2}, \infty\right) \\[5ex]$ For the range, there is only the plus sign in front of the radical.
The smallest "real number" output value occurs for the smallest input value.
The smallest input value is $-\dfrac{5}{2}$

$5 + 2x \\[3ex] 5 + 2 * -\dfrac{5}{2} \\[5ex] 5 - 5 \\[3ex] 0 \\[3ex] \sqrt{0} = 0 \\[3ex]$ The smallest output value is $0$
The range can only include nonnegative numbers (zero and positive numbers).

$R = \{y | y \ge 0\} \\[3ex] R = [0, \infty) \\[3ex]$ (b.)
It is a function (positive square root function). It is also a one-to-one function.
Each input has exactly one output.

To find the domain, the radicand (the value inside the square root radical) must not be negative.
Square root of negative numbers are not real numbers. They are imaginary numbers.
We are only dealing with real numbers.
So, the radicand must be a nonnegative number (zero and positive number).
That means that the radicand must be at least zero.

$2x - 5 \ge 0 \\[3ex] 2x \ge 0 + 5 \\[3ex] 2x \ge 5 \\[3ex] x \ge \dfrac{5}{2} \\[5ex] D = \{x : x \ge \dfrac{5}{2}\} \\[5ex] D = \left[\dfrac{5}{2}, \infty\right) \\[5ex]$ For the range, there is only the plus sign in front of the radical.
The smallest "real number" output value occurs for the smallest input value.
The smallest input value is $\dfrac{5}{2}$

$2 * \dfrac{5}{2} - 5 \\[5ex] 5 - 5 = 0 \\[3ex] \sqrt{0} = 0 \\[3ex]$ The smallest output value is $0$
The range can only include nonnegative numbers (zero and positive numbers).

$R = \{y | y \ge 0\} \\[3ex] R = [0, \infty) \\[3ex]$ (c.)

$g(x) = \sqrt{3x - 4} \\[3ex] h(x) = 3x + 5 \\[3ex] g(h(x)) \\[3ex] = g(3x + 5) \\[3ex] = \sqrt{3(3x + 5) - 4} \\[3ex] = \sqrt{9x + 15 - 4} \\[3ex] = \sqrt{9x + 11} \\[3ex]$ The composite (the result of the composition of the functions) is a function (positive square root function).
It is also a one-to-one function.
Each input has an exactly one output.

To find the domain, the radicand (the value inside the square root radical) must not be negative.
Square root of negative numbers are not real numbers. They are imaginary numbers.
We are only dealing with real numbers.
So, the radicand must be a nonnegative number (zero and positive number).
That means that the radicand must be at least zero.

$9x + 11 \ge 0 \\[3ex] 9x \ge 0 - 11 \\[3ex] 9x \ge -11 \\[3ex] x \ge -\dfrac{11}{9} \\[5ex] D = \{x : x \ge -\dfrac{11}{9}\} \\[5ex] D = \left[-\dfrac{11}{9}, \infty\right) \\[5ex]$ For the range, there is only the plus sign in front of the radical.
The smallest "real number" output value occurs for the smallest input value.
The smallest input value is $-\dfrac{11}{9}$

$9x + 11 \\[3ex] 9 * -\dfrac{11}{9} + 11 \\[5ex] -11 + 11 \\[3ex] 0 \\[3ex] \sqrt{0} = 0 \\[3ex]$ The smallest output value is $0$
The range can only include nonnegative numbers (zero and positive numbers).

$R = \{y | y \ge 0\} \\[3ex] R = [0, \infty)$
(48.) Please visit: Domain and Range from Graphs: Example 6 (https://openstax.org/books/college-algebra/pages/3-2-domain-and-range)
Scroll down till you see: Finding Domain and Range from Graphs

(a.) Give students some input values for each relation and ask them to calculate the output values
What will be the output values for those input values?

(b.) Give students some output values and ask them to calculate the input values.
What input values will give those output values?

(49.) Detemine if the following statement is true or false.
The domain of $\dfrac{f(x)}{g(x)}$ consists of numbers x that are in the domains of both f and g

A. ​True, because​ f(x) and​ g(x) exist for any x in the domain of both f and g.
B. ​False, because​ g(x) may equal zero for some x values in the domain of g.
C. ​True, because $\dfrac{f(x)}{g(x)}$ is always defined for any x in the domain of both f and g.
D. ​False, because​ f(x) may equal zero for some x values in the domain of f.

For the function: $\dfrac{f(x)}{g(x)}$, what if $g(x)$ is zero?

For the function: $\dfrac{f(x)}{g(x)}$:
if $g(x)$ = 0, then the function: $\dfrac{f(x)}{g(x)}$ is undefined because the denominator cannot be zero (division by zero)
This means that the values of x for which $g(x)$ is zero, will make the function undefined.
This means that the function: $\dfrac{f(x)}{g(x)}$ is not defined on any value of x for which $g(x)$ is zero
This means that the function: $\dfrac{f(x)}{g(x)}$ will not contain those values of x in it's domain
Because it is possible that $g(x)$ may contain those values of x (that will make it zero) in it's domain, we cannot definitely say that the function: $\dfrac{f(x)}{g(x)}$ domain of $\dfrac{f(x)}{g(x)}$ consists of numbers x in both domains of f and g
Therefore, the answer is: B. ​False, because​ g(x) may equal zero for some x values in the domain of g.

The domain of $\dfrac{f(x)}{g(x)}$ consists of all real numbers x for which ​$g(x) \ne 0$ that are also in the domain of both f and g.
(50.) Please visit: Domain and Range from Graphs: Example 7 (https://openstax.org/books/college-algebra/pages/3-2-domain-and-range)
Scroll down till you see: Finding Domain and Range from Graphs

(a.) Give students some input values for each relation and ask them to calculate the output values
What will be the output values for those input values?

(b.) Give students some output values and ask them to calculate the input values.
What input values will give those output values?

(51.) $y = \sqrt[3]{x}$
The same answer applies to $y = \sqrt[5]{x}$, $y = \sqrt[7]{x}$, $y = \sqrt[9]{x}$, ...
Ask students to explain why the functions with odd-th roots (odd-numbered roots) have the same domain and range.

It is a function (positive cube root function). It is also a one-to-one function.
Each input has exactly one output.

The cube root of a positive number is a positive number.
The cube root of zero is zero.
The cube root of a negative number is a negative number.
So, the radicand can be any number.

$D = \{x | x \in \mathbb{R}\} \\[3ex] D = (-\infty, \infty) \\[3ex]$ Similarly, the range is:

$R = \{y | y \in \mathbb{R}\} \\[3ex] R = (-\infty, \infty)$
(52.) Domain and Range from Graphs: Try It #6
Please scroll till you see: Finding Domain and Range from Graphs

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