# Solved Examples on Functions (all topics)

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, CMAT, and NZQA Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

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.

Solve all questions.
Show all work.
Use at least two methods to solve the questions as applicable.

(1.)

(2.) If x = −4, determine the value of $3x^2 - 5x + \dfrac{1}{x}$

$3x^2 - 5x + \dfrac{1}{x} \\[5ex] x = -4 \\[3ex] \implies \\[3ex] 3(-4)^2 - 5(-4) + \dfrac{1}{-4} \\[5ex] 3(16) + 20 - \dfrac{1}{4} \\[5ex] 48 + 20 - \dfrac{1}{4} \\[5ex] 68 - \dfrac{1}{4} \\[5ex] \dfrac{272}{4} - \dfrac{1}{4} \\[5ex] \dfrac{272 - 1}{4} \\[5ex] \dfrac{271}{4} \\[5ex] 67.75$
(3.) Let P = (x, y) be a point on the graph of y = x² − 4
(a.) Express the distance d from P to the origin as a function of x

(b.) What is d if x = 0?
Round to two decimal places as needed.

(c.) What is d if x = 1?
Round to two decimal places as needed.

(d.) Use a graphing utility to graph d = d(x).
For what positive value of x is d smallest?
Round to two decimal places as needed.

$(a.) \\[3ex] From\;\;P(x, y) \;\;to\;\;Origin(0, 0) \\[3ex] x_1 = x \\[3ex] y_1 = y \\[3ex] x_2 = 0 \\[3ex] y_2 = 0 \\[3ex] d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[3ex] = \sqrt{(0 - x)^2 + (0 - y)^2} \\[3ex] = \sqrt{(-x)^2 + (-y)^2} \\[3ex] = \sqrt{x^2 + y^2} \\[3ex] As\;\;a\;\;function\;\;of\;\;x: \\[3ex] y = x^2 - 4 \\[3ex] y^2 = (x^2 - 4)^2 \\[3ex] y^2 = (x^2 - 4)(x^2 - 4) \\[3ex] y^2 = x^4 - 4x^2 - 4x^2 + 16 \\[3ex] y^2 = x^4 - 8x^2 + 16 \\[3ex] \implies \\[3ex] d = \sqrt{x^2 + x^4 - 8x^2 + 16} \\[3ex] d(x) = \sqrt{x^4 - 7x^2 + 16} \\[5ex] (b.) \\[3ex] x = 0 \\[3ex] d(0) = \sqrt{0^4 - 7(0)^2 + 16} \\[3ex] = \sqrt{0 - 7(0) + 16} \\[3ex] = \sqrt{0 - 0 + 16} \\[3ex] = \sqrt{16} \\[3ex] = 4 \\[5ex] (c.) \\[3ex] x = 1 \\[3ex] d(1) = \sqrt{1^4 - 7(1)^2 + 16} \\[3ex] = \sqrt{1 - 7(1) + 16} \\[3ex] = \sqrt{1 - 7 + 16} \\[3ex] = \sqrt{10} \\[3ex] = 3.16227766 \\[3ex] \approx 3.16 \\[5ex]$ (d.) The graph of $x^4 - 7x^2 + 16$ is:

The positive value of x when d is smallest is 1.870828693387 ≈ 1.87
(4.) ACT For functions $f(x) = 5 \cdot 2^x$ and $g(x) = 10x$, the value of f(3) − g(3) is:

$A.\;\; 0 \\[3ex] B.\;\; 10 \\[3ex] C.\;\; 70 \\[3ex] D.\;\; 970 \\[3ex] E.\;\; 1,030 \\[3ex]$

$f(x) = 5 \cdot 2^x \\[3ex] f(3) = 5 \cdot 2^3 \\[3ex] f(3) = 5 \cdot 8 \\[3ex] f(3) = 40 \\[5ex] g(x) = 10x \\[3ex] g(3) = 10(3) \\[3ex] g(3) = 30 \\[5ex] f(3) - g(3) \\[3ex] = 40 - 30 \\[3ex] = 10$
(5.)

(6.)

(7.) ACT Given the sets A = {0, 1, 2, 3} and B = {1, 3, 5, 7}. which of the following defines a function f from A onto B?

$F.\;\; f(x) = 2x + 1 \\[3ex] G.\;\; f(x) = 2x - 1 \\[3ex] H.\;\; f(x) = 3x - 1 \\[3ex] J.\;\; f(x) = 3x - 2 \\[3ex] K.\;\; f(x) = x + 1 \\[3ex]$

We can do this question in at least two ways.
Use any approach you prefer.

$\underline{First\;\;Approach:\;\;Testing} \\[3ex] Recommended\;\;for\;\;ACT \\[3ex] Mapping\;\;from\;\;A\;\;onto\;\;B \\[3ex] F.\;\; f(x) = 2x + 1 \\[3ex] f(0) = 2(0) + 1 = 0 + 1 = 1 ...okay \\[3ex] f(1) = 2(1) + 1 = 2 + 1 = 3 ...okay \\[3ex] f(2) = 2(2) + 1 = 4 + 1 = 5 ...okay \\[3ex] f(3) = 2(3) + 1 = 6 + 1 = 7 ...okay \\[3ex]$ Option F. is the answer.

$\underline{First\;\;Approach:\;\;Calculating} \\[3ex] Mapping\;\;from\;\;A\;\;onto\;\;B \\[3ex] Use\;\;any\;\;two\;\;points\;\;and\;\;the\;\;corresponding\;\;points \\[3ex] A = \{0, 1, 2, 3\} \\[3ex] x_1 = 1 \\[3ex] x_2 = 2 \\[3ex] B = \{1, 3, 5, 7\} \\[3ex] y_1 = 3 \\[3ex] y_2 = 5 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{5 - 3}{2 - 1} \\[5ex] = \dfrac{2}{1} \\[5ex] = 2 \\[3ex] y = mx + b \\[3ex] For\;\; (3, 7) \\[3ex] x = 3 \\[3ex] y = 7 \\[3ex] 7 = 2(3) + b \\[3ex] 7 = 6 + b \\[3ex] 6 + b = 7 \\[3ex] b = 7 - 6 \\[3ex] b = 1 \\[3ex] \implies \\[3ex] y = 2x + 1 \\[3ex] f(x) = 2x + 1$
(8.)

(9.) CSEC The function $f$ is defined as $f : x \rightarrow 3 - 2x$
(i.) The diagram below shows the mapping diagram of the function, $f$
Determine the value of $a$

(ii.) Determine, in their simplest form, expressions for
(a.) the inverse of the function $f$, $f^{-1}(x)$
(b.) the composite function $f^2(x)$

(iii.) State the value of $ff^{-1}(-2)$

$(i) \\[3ex] x = -1 \\[3ex] f(-1) = a...Mapping\;\;Diagram \\[3ex] f : x \rightarrow 3 - 2x \\[3ex] f(x) = 3 - 2x \\[3ex] f(-1) = 3 - 2(-1) \implies a = 3 - 2(-1) \\[3ex] a = 3 + 2 \\[3ex] a = 5 \\[3ex] (ii) \\[3ex] (a.) \\[3ex] f(x) = 3 - 2x \\[3ex] Let\;\; y = 3 - 2x \\[3ex] Interchange\;\;x\;\;and\;\;y \\[3ex] x = 3 - 2y \\[3ex] Solve\;\;for\;\;y \\[3ex] x + 2y = 3 \\[3ex] 2y = 3 - x \\[3ex] y = \dfrac{3 - x}{2} \\[5ex] \therefore f^{-1}(x) = \dfrac{3 - x}{2} \\[5ex] (b.) \\[3ex] Composite\;\;Function:\;\; f^2(x) = (f \circ f)(x) \\[3ex] = f(f(x)) \\[3ex] = f(3 - 2x) \\[3ex] = 3 - 2(3 - 2x) \\[3ex] = 3 - 6 + 4x \\[3ex] = 4x - 3 \\[3ex] (iii) \\[3ex] ff^{-1}(-2) \\[3ex] Start\;\;from\;\;inside \\[3ex] f^{-1}(x) = \dfrac{3 - x}{2} \\[5ex] f^{-1}(-2) \\[3ex] = \dfrac{3 - -2}{2} \\[5ex] = \dfrac{3 + 2}{2} \\[5ex] = \dfrac{5}{2} \\[5ex] Then\;\;to\;\;outside \\[3ex] ff^{-1}(-2) \\[3ex] = f\left(\dfrac{5}{2}\right) \\[5ex] = 3 - 2\left(\dfrac{5}{2}\right) \\[5ex] = 3 - 5 \\[3ex] = -2$
(10.)

$f(-a)$
This means that we have to substitute $(-a)$ for $x$

$f(x) = x^2 - 3x \\[3ex] f(-a) = (-a)^2 - 3(-a) \\[3ex] = a^2 + 3a \\[3ex] = a(a + 3)$

$f(a - 3)$
This means that we have to substitute $(a - 3)$ for $x$

$f(x) = x^2 - 3x \\[3ex] f(a - 3) = (a - 3)^2 - 3(a - 3) \\[3ex] = (a - 3)(a - 3) - 3a + 9 \\[3ex] = a^2 - 3a - 3a + 9 - 3a + 9 \\[3ex] = a^2 - 9a + 18 \\[3ex] = (a - 3)(a - 6)$
OR
$f(x) = x^2 - 3x \\[3ex] f(a - 3) = (a - 3)^2 - 3(a - 3) \\[3ex] = (a - 3)(a - 3) - 3a + 9 \\[3ex] = (a - 3)(a - 3) - 3(a - 3) \\[3ex] = (a - 3)[(a - 3) - 3] \\[3ex] = (a - 3)[a - 3 - 3] \\[3ex] = (a - 3)(a -6)$

$f(a + h)$
This means that we have to substitute $(a + h)$ for $x$

$f(x) = x^2 - 3x \\[3ex] f(a + h) = (a + h)^2 - 3(a + h) \\[3ex] = (a + h)(a + h) - 3(a + h) \\[3ex] = (a + h)[(a + h) - 3] \\[3ex] = (a + h)(a + h - 3)$
(11.) CSEC (a) The functions f(x) and g(x) are defined as:

$f(x) = \dfrac{5x - 4}{3} \\[5ex] g(x) = x^2 - 1 \\[3ex]$ (i) Evaluate $f(7)$
(ii) Write an expression, in terms of $x$, for $f^{-1}(x)$
(iii) Write an expression, in terms of $x$, for $fg(x)$

(b) (i) Express the quadratic function $f(x) = 3x^2 + 6x - 2$, in the form $a(x + h)^2 + k$, where $a, h$ and $k$ are constants.
(ii) Hence, or otherwise, state the minimum value of $f(x) = 3x^2 + 6x -2$
(iii) State the equation of the axis of symmetry of the function $f(x) = 3x^2 + 6x -2$
(iv) Sketch the graph of $y = 3x^2 + 6x - 2$, showing on your sketch
(a) the intercept on the $y-axis$
(b) the coordinates of the minimum point.

$(a) \\[3ex] (i) \\[3ex] f(x) = \dfrac{5x - 4}{3} \\[5ex] f(7) = \dfrac{5(7) - 4}{3} \\[5ex] f(7) = \dfrac{35 - 4}{3} \\[3ex] f(7) = \dfrac{31}{3} \\[5ex] (ii) \\[3ex] Equate\:\:y = f(x) \\[3ex] But\:\:Let\:\:y = f^{-1}(x) \\[3ex] y = \dfrac{5x - 4}{3} \\[5ex] Interchange\:\:x\:\:and\:\:y \\[3ex] x = \dfrac{5y - 4}{3} \\[5ex] Solve\:\:for\:\:y \\[3ex] \dfrac{5y - 4}{3} = x \\[5ex] 5y - 4 = 3x \\[3ex] 5y = 3x + 4 \\[3ex] y = \dfrac{3x + 4}{5} \\[5ex] \therefore f^{-1}(x) = \dfrac{3x + 4}{5} \\[5ex] (iii) \\[3ex] fg(x) = (fg)(x) = f(x) * g(x) \\[3ex] (fg)(x) = \dfrac{5x - 4}{3} * x^2 - 1 \\[5ex] (fg)(x) = \dfrac{(5x - 4)(x^2 - 1)}{3} \\[5ex] x^2 - 1 = x^2 - 1^2 = (x + 1)(x - 1)...Difference\:\:of\:\:Two\:\:Squares \\[3ex] (fg)(x) = \dfrac{(5x - 4)(x + 1)(x - 1)}{3} \\[5ex] (b) \\[3ex] (i) \\[3ex] Standard\;\;Form:\;\;f(x) = 3x^2 + 6x - 2 \\[3ex] Compare:\;\; f(x) = ax^2 + bx + c \\[3ex] \implies a = 3,\;\;b = 6,\;\;c = -2 \\[3ex] \underline{1st\;\;Method:\;\;Vertex\;\;Formula} \\[3ex] (h, k) = \left(-\dfrac{b}{2a}, f\left(-\dfrac{b}{2a}\right)\right) \\[5ex] h = -\dfrac{b}{2a} \\[5ex] h = -\dfrac{6}{2(3)} \\[5ex] h = -1 \\[3ex] k = f\left(-\dfrac{b}{2a}\right) \\[5ex] k = f(-1) \\[3ex] k = 3(-1)^2 + 6(-1) - 2 \\[3ex] k = 3(1) - 6 - 2 \\[3ex] k = 3 - 6 - 2 \\[3ex] k = -5 \\[3ex] (h, k) = (-1, -5) \\[3ex] Vertex\;\;Form:\;\; f(x) = a(x - h)^2 + k \\[3ex] Vertex\;\;Form:\;\; f(x) = 3(x - -1)^2 + -5 \\[3ex] Vertex\;\;Form:\;\; f(x) = 3(x + 1)^2 - 5 \\[3ex] \underline{2nd\;\;Method:\;\;Completing\;\;the\;\;Square\;\;Method} \\[3ex] Standard\;\;Form:\;\;f(x) = 3x^2 + 6x - 2 \\[3ex] f(x) = 3\left(x^2 + \dfrac{6x}{3} - \dfrac{2}{3}\right) \\[5ex] f(x) = 3\left(x^2 + 2x - \dfrac{2}{3}\right) \\[5ex] Coefficient\;\;of\;\;x = 2 \\[3ex] Half\;\;of\;\;it = \dfrac{1}{2} * 2 = 1 \\[5ex] Square\;\;it = (1)^2 \\[3ex] f(x) = 3\left[x^2 + 2x + (1)^2 - \dfrac{2}{3} - (1)^2\right] \\[5ex] f(x) = 3\left[(x + 1)^2 - \dfrac{2}{3} - 1\right] \\[5ex] f(x) = 3\left[(x + 1)^2 - \dfrac{2}{3} - \dfrac{3}{3}\right] \\[5ex] f(x) = 3\left[(x + 1)^2 - \dfrac{5}{3}\right] \\[5ex] f(x) = 3(x + 1)^2 - 5 \\[5ex] (ii) \\[3ex] Minimum\;\;value = y-coordinate\;\;of\;\;the\;\;Vertex \\[3ex] Vertex = (h, k) = (-1, -5) \\[3ex] Minimum\;\;value = k = -5 \\[3ex] (iii) \\[3ex] Axis\;\;of\;\;symmetry = x-coordinate\;\;of\;\;the\;\;Vertex \\[3ex] Vertex = (h, k) = (-1, -5) \\[3ex] Axis\;\;of\;\;symmetry = h = -1 \\[3ex] Equation\;\;of\;\;the\;\;axis\;\;of\;\;symmetry:\;\; x = -1 \\[3ex] (iv) \\[3ex] (a.) \\[3ex] To\;\;find\;\;y-intercept \\[3ex] Set\;\;x = 0\;\;and\;\;solve\;\;for\;\;y \\[3ex] y = 3x^2 + 6x - 2 \\[3ex] y = 3(0)^2 + 6(0) - 2 \\[3ex] y = 3(0) + 0 - 2 \\[3ex] y = 0 + 0 - 2 \\[3ex] y = -2 \\[3ex] y-intercept = (0, -2) \\[3ex] (b.) \\[3ex] Minimum\;\;point = Vertex = (h, k) = (-1, -5) \\[3ex]$
(12.) WASSCE The functions f and g are defined as

$f:x \rightarrow 2 - x^2 \:\:and \\[3ex] g:x \rightarrow \dfrac{1}{x - 1} \\[5ex]$ Evaluate

$(i)\:\: g\left(-\dfrac{1}{4}\right) \\[5ex] (ii)\:\: \dfrac{f(2)}{g(3)} \\[5ex]$

$(i) \\[3ex] g:x \rightarrow \dfrac{1}{x - 1} \\[5ex] g\left(-\dfrac{1}{4}\right) \implies x = \dfrac{1}{4} \\[5ex] g\left(-\dfrac{1}{4}\right) = \dfrac{1}{-\dfrac{1}{4} - 1} \\[7ex] -\dfrac{1}{4} - 1 = \dfrac{-1}{4} - \dfrac{4}{4} = \dfrac{-1 - 4}{4} = -\dfrac{5}{4} \\[5ex] \dfrac{1}{-\dfrac{1}{4} - 1} = \dfrac{1}{-\dfrac{5}{4}} = 1 \div -\dfrac{5}{4} = 1 * -\dfrac{4}{5} = -\dfrac{4}{5} \\[7ex] \therefore g\left(-\dfrac{1}{4}\right) = -\dfrac{4}{5} \\[5ex] (ii) \\[3ex] f:x \rightarrow 2 - x^2 \\[3ex] f(2) \implies x = 2 \\[3ex] f(2) = 2 - (2)^2 \\[3ex] f(2) = 2 - 4 \\[3ex] f(2) = -2 \\[3ex] g:x \rightarrow \dfrac{1}{x - 1} \\[5ex] g(3) \implies x = 3 \\[3ex] g(3) = \dfrac{1}{3 - 1} \\[5ex] g(3) = \dfrac{1}{2} \\[5ex] \dfrac{f(2)}{g(3)} = f(2) \div g(3) \\[5ex] \dfrac{f(2)}{g(3)} = -2 \div \dfrac{1}{2} = -2 * \dfrac{2}{1} = -2 * 2 = -4 \\[5ex] \therefore \dfrac{f(2)}{g(3)} = -4$
(13.) CSEC
(a.) Given that $f(x) = 4x - 7$ and $g(x) = \dfrac{3x + 1}{2}$, determine the values of:

$(i)\;\; g(0) + g(5) \\[3ex] (ii)\;\; fg(5) \\[3ex] (iii)\;\; f^{-1}(1) \\[3ex]$ (b) P(6, -1) and Q(2, 7) are the end points of a line segment PQ
Determine
(ii) the coordinates of the midpoint of PQ
(iii) the equation of the perpendicular bisector of PQ

$(a) \\[3ex] (i) \\[3ex] g(x) = \dfrac{3x + 1}{2} \\[5ex] g(0) = \dfrac{3(0) + 1}{2} = \dfrac{0 + 1}{2} = \dfrac{1}{2} \\[5ex] g(5) = \dfrac{3(5) + 1}{2} = \dfrac{15 + 1}{2} = \dfrac{16}{2} = 8 \\[5ex] g(0) + g(5) \\[3ex] = \dfrac{1}{2} + 8 \\[5ex] = \dfrac{1}{2} + \dfrac{16}{2} \\[5ex] = \dfrac{1 + 16}{2} \\[5ex] = \dfrac{17}{2} \\[5ex] (ii) \\[3ex] Two\;\;methods\;\;to\;\;solve\;\;this\;\;question \\[3ex] Use\;\;whichever\;\;method\;\;you\;\;prefer \\[3ex] f(x) = 4x - 7 \\[3ex] g(x) = \dfrac{3x + 1}{2} \\[5ex] \underline{1st\;\;Method} \\[3ex] fg(5) = f(5) * g(5) \\[3ex] f(x) = 4x - 7 \\[3ex] f(5) = 4(5) - 7 = 20 - 7 = 13 \\[3ex] g(5) = 8...from\;\;(i) \\[3ex] \therefore fg(5) = 13(8) = 104 \\[3ex] \underline{2nd\;\;Method} \\[3ex] fg(x) \\[3ex] = f(x) * g(x) \\[3ex] = (4x - 7) * \dfrac{3x + 1}{2} \\[5ex] = \dfrac{(4x - 7)(3x + 1)}{2} \\[5ex] fg(5) \\[3ex] = \dfrac{[4(5) - 7][3(5) + 1]}{2} \\[5ex] = \dfrac{(20 - 7)(15 + 1)}{2} \\[5ex] = \dfrac{13(16)}{2} \\[5ex] = 13(8) \\[3ex] = 104 \\[3ex] (iii) \\[3ex] f(x) = 4x - 7 \\[3ex] Let\;\;y = 4x - 7 \\[3ex] Interchange\;\;x\;\;and\;\;y \\[3ex] x = 4y - 7 \\[3ex] Solve\;\;for\;\;y \\[3ex] Swap \\[3ex] 4y - 7 = x \\[3ex] 4y = x + 7 \\[3ex] y = \dfrac{x + 7}{4} \\[5ex] \therefore f^{-1}(x) = \dfrac{x + 7}{4} \\[5ex] f^{-1}(1) \\[3ex] = \dfrac{1 + 7}{4} \\[5ex] = \dfrac{8}{4} \\[5ex] = 2 \\[3ex] (b) \\[3ex] P(6, -1) \rightarrow x_1 = 6, \;\;\; y_1 = -1 \\[3ex] Q(2, 7) \rightarrow x_2 = 2, \;\;\; y_2 = 7 \\[3ex] (i) \\[3ex] m = Gradient\;\;of\;\;PQ \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{7 - (-1)}{2 - 6} \\[5ex] m = \dfrac{7 + 1}{-4} \\[5ex] m = \dfrac{8}{-4} \\[5ex] m = -2 \\[3ex] (ii) \\[3ex] Midpoint\;\;of\;\;PQ = \overline{PQ} \\[3ex] \overline{PQ} = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right) \\[5ex] = \left(\dfrac{6 + 2}{2}, \dfrac{-1 + 7}{2}\right) \\[5ex] = \left(\dfrac{8}{2}, \dfrac{6}{2}\right) \\[5ex] = (4, 3) \\[3ex] (iii) \\[3ex] Midpoint = Point = (4,3) \rightarrow x_1 = 4, \;\;\; y_1 = 3 \\[3ex] Gradient = m = -2 \\[3ex] Gradient\;\;of\;\;Perpendicular\;\;Line = -\dfrac{1}{m} = \dfrac{-1}{-2} = \dfrac{1}{2} \\[5ex] Point-Slope\;\;Form \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 3 = \dfrac{1}{2}(x - 4) \\[5ex] y - 3 = \dfrac{1}{2}x - 2 \\[5ex] y = \dfrac{1}{2}x - 2 + 3 \\[5ex] y = \dfrac{1}{2}x + 1$
(14.) ACT If $f(x) = x^2 + x + 5$ and $g(x) = \sqrt{x}$,

then what is the value of $\dfrac{g(4)}{f(1)}$

$A.\:\: \dfrac{2}{7} \\[5ex] B.\:\: \dfrac{25}{7} \\[5ex] C.\:\: \dfrac{2}{25} \\[5ex] D.\:\: 2 \\[3ex] E.\:\: 4 \\[3ex]$

$f(x) = x^2 + x + 5 \\[3ex] f(1) = 1^2 + 1 + 5 \\[3ex] f(1) = 1 + 1 + 5 \\[3ex] f(1) = 7 \\[3ex] g(x) = \sqrt{x} \\[3ex] g(4) = \sqrt{4} \\[3ex] g(4) = 2 \\[3ex] \dfrac{g(4)}{f(1)} = \dfrac{2}{7}$
(15.) CSEC The functions f(x) and g(x) are defined as

$f(x) = 3x + 2 \hspace{7em} g(x) = \dfrac{x^2 - 1}{3} \\[5ex]$ (i) Evaluate $g(5)$

(ii) Write an expression in terms of $x$ for $f^{-1}(x)$

(iii) Write an expression for $g(f(x))$, in the form $(x + a)(x + b)$, where $a, b \in R$

$(i) \\[3ex] g(x) = \dfrac{x^2 - 1}{3} \\[5ex] g(5) \implies x = 5 \\[3ex] g(5) = \dfrac{5^2 - 1}{3} \\[5ex] g(5) = \dfrac{25 - 1}{3} \\[5ex] g(5) = \dfrac{24}{3} \\[5ex] g(5) = 8 \\[3ex] (ii) \\[3ex] f(x) = 3x + 2 \\[3ex] Let\:\: y = 3x + 2 \\[3ex] Interchange\:\:x\:\:and\:\:y \\[3ex] x = 3y + 2 \\[3ex] Solve\:\:for\:\:y \\[3ex] 3y + 2 = x \\[3ex] 3y = x - 2 \\[3ex] y = \dfrac{x - 2}{3} \\[5ex] New\:\:y\:\:becomes\:\:f^{-1}(x) \\[3ex] \therefore f^{-1}(x) = \dfrac{x - 2}{3} \\[5ex] (iii) \\[3ex] g(f(x)) = g(3x + 2) \\[3ex] g(x) = \dfrac{x^2 - 1}{3} \\[5ex] g(3x + 2) \implies x = 3x + 2 \\[3ex] g(3x + 2) = \dfrac{(3x + 2)^2 - 1}{3} \\[5ex] (3x + 2)^2 - 1 = (3x + 2)^2 - 1^2 \\[3ex] (3x + 2)^2 - 1^2 = [(3x + 2) + 1][(3x + 2) - 1]...Difference\:\:of\:\:Two\:\:Squares \\[3ex] = (3x + 2 + 1)(3x + 2 - 1) \\[3ex] = (3x + 3)(3x + 1) \\[3ex] = 3(x + 1)(3x + 1) \\[3ex] \rightarrow \dfrac{(3x + 2)^2 - 1}{3} = \dfrac{3(x + 1)(3x + 1)}{3} \\[5ex] = (x + 1)(3x + 1) \\[3ex] \rightarrow g(3x + 2) = (x + 1)(3x + 1) \\[3ex] \therefore g(f(x)) = (x + 1)(3x + 1)$
(16.) ACT For functions $f(x) = 5 * 2^x$ and $g(x) = 10x$, the value of f(3) − g(3) is:

$A.\:\: 0 \\[3ex] B.\:\: 10 \\[3ex] C.\:\: 70 \\[3ex] D.\:\: 970 \\[3ex] E.\:\: 1,030 \\[3ex]$

$f(x) = 5 * 2^x \\[3ex] f(3) = 5 * 2^3 \\[3ex] f(3) = 5 * 8 \\[3ex] f(3) = 40 \\[3ex] g(x) = 10x \\[3ex] g(3) = 10(3) \\[3ex] g(3) = 30 \\[3ex] f(3) - g(3) \\[3ex] = 40 - 30 \\[3ex] = 10$
(17.) A cylinder (round can) has a circular base and a circular top with vertical sides in between.
Let $r$ be the radius of the top of the can and let $h$ be the height.
The surface area of the cylinder, $A$, is $A = 2\pi r^2 + 2\pi rh$ (it's two circles for the top and bottom plus a rolled up rectangle for the side).

Assume that the height of the cylinder is $12$ inches.

(a.) Determine the domain of $A(r)$
In other words, determine the values of $r$ for which $A(r)$ is defined.

(b.) Write the radius as a function of $A$
In other words, determine the inverse function of $A(r)$

(c.) If the surface area is 300 square inches, what is the radius?
In other words, evaluate $r(300)$

$A = 2\pi r^2 + 2\pi rh \\[3ex] h = 12\;inches \\[3ex] A = 2\pi r^2 + 2 * \pi * r * 12 \\[3ex] A = 2\pi r^2 + 24\pi r \\[3ex]$ To determine the domain:
The area cannot be non-negative
It cannot be zero or negative.
The area must be positive (greater than zero)
Similarly, the radius must be positive (greater than zero)

$A = 2\pi r^2 + 24\pi r \\[3ex] 2\pi r^2 + 24\pi r = A \\[3ex] A \gt 0 \\[3ex] \implies 2\pi r^2 + 24\pi r \gt 0 \\[3ex] 2\pi(r^2 + 12r) \gt 0 \\[3ex] r^2 + 12r \gt \dfrac{0}{2\pi} \\[5ex] r^2 + 12r \gt 0 \\[3ex] r(r + 12) \gt 0 \\[3ex] \underline{Boundary\;\;Values} \\[3ex] r = 0 \\[3ex] r + 12 = 0 \\[3ex] r = -12 \\[3ex] \underline{Test\;\;Intervals} \\[3ex] r \lt -12 \\[3ex] -12 \lt r \lt 0 \\[3ex] r \gt 0 \\[3ex]$
 Test Intervals: $r \lt -12$ $-12 \lt r \lt 0$ $r \gt 0$ $r$ $-$ $-$ $+$ $r + 12$ $-$ $+$ $+$ $r(r + 12)$ $+$ $-$ $+$

$r(r + 12) \gt 0 \\[3ex] r \lt -12 \cup r \gt 0 \\[3ex]$ Let us check the solution of the Quadratic Inequality to ensure we are correct

 LHS RHS $r(r + 12)$ $0$ $(\infty, -12)$ Let $r = -16$ $r(r + 12)$ $-16(-16 + 12)$ $-16(-4)$ $64$ $0$ $64 \gt 0$ $(0, \infty)$ Let $r = 1$ $1(1 + 12)$ $1(13)$ $13$ $0$ $13 \gt 0$

$(a.) \\[3ex] Because\;\;r\;\;must\;\;be\;\;positive \\[3ex] \underline{Solution:\;\;Interval\;\;Notation} \\[3ex] Domain:\;\; (0, \infty) \\[3ex]$ Student: Mr. C, I have two questions.
First one: do we need to go through all these steps?
So, after writing the test intervals; we can simply write that the domain is $r \gt 0$
We do not need to actually test it and check it.
Is that correct?
Teacher: That is correct
Student: Okay. The next question is
Do we have to write it in interval notation?
Teacher: Because the domain is a set, it is important we express it in set notation and/or interval notation.

Set Notation: $\{r: r \gt 0\}$

Interval Notation: $(0, \infty)$

$(b.) \\[3ex] A = 2\pi r^2 + 24\pi r \\[3ex] 2\pi r^2 + 24\pi r = A \\[3ex] 2\pi r^2 + 24\pi r - A = 0 \\[3ex] This\;\;is\;\;quadratic\;\;in\;\; r \\[3ex] We\;\;can\;\;solve\;\;it\;\;in\;\;at\;\;least\;\;two\;\;ways \\[3ex] \underline{First\;\;Method:\;\;Quadratic\;\;Formula} \\[3ex] a = 2\pi \\[3ex] b = 24\pi \\[3ex] c = -A \\[3ex] r = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] r = \dfrac{-24\pi \pm \sqrt{(24\pi)^2 - 4(2\pi)(-A)}}{2(2\pi)} \\[5ex] r = \dfrac{-24\pi \pm \sqrt{576\pi^2 + 8\pi A}}{4\pi} \\[5ex] r = \dfrac{-24\pi \pm \sqrt{4(144\pi^2 + 2\pi A)}}{4\pi} \\[5ex] r = \dfrac{-24\pi \pm 2\sqrt{144\pi^2 + 2\pi A}}{4\pi} \\[5ex] r = \dfrac{2\left(-12\pi \pm \sqrt{144\pi^2 + 2\pi A}\right)}{4\pi} \\[5ex] r = \dfrac{-12\pi \pm \sqrt{144\pi^2 + 2\pi A}}{2\pi} \\[5ex] r = \dfrac{-12\pi \pm \sqrt{2\pi(72\pi + A)}}{2\pi} \\[7ex] \underline{Second\;\;Method:\;\;Completing\;\;the\;\;Square\;\;Method} \\[3ex] 2\pi r^2 + 24\pi r = A \\[3ex] \dfrac{2\pi r^2}{2\pi} + \dfrac{24\pi r}{2\pi} = \dfrac{A}{2\pi} \\[5ex] r^2 + 12r = \dfrac{A}{2\pi} \\[5ex] Coefficient\;\;of\;\;r = 12 \\[3ex] Half\;\;of\;\;it = \dfrac{12}{2} = 6 \\[5ex] Square\;\;it = 6^2 \\[3ex] r^2 + 12r + 6^2 = \dfrac{A}{2\pi} + 6^2 \\[5ex] (r + 6)^2 = \dfrac{A}{2\pi} + 36 \\[5ex] (r + 6)^2 = \dfrac{A + 72\pi}{2\pi} \\[5ex] r + 6 = \pm \sqrt{\dfrac{A + 72\pi}{2\pi}} \\[5ex] r + 6 = \pm \dfrac{\sqrt{A + 72\pi}}{\sqrt{2\pi}} \\[5ex] r + 6 = \pm \dfrac{\sqrt{A + 72\pi}}{\sqrt{2\pi}} * \dfrac{\sqrt{2\pi}}{\sqrt{2\pi}} \\[5ex] r + 6 = \pm \dfrac{\sqrt{2\pi(A + 72\pi)}}{2\pi} \\[5ex] r = -6 \pm \dfrac{\sqrt{2\pi(A + 72\pi)}}{2\pi} \\[5ex] r = -\dfrac{12\pi}{2\pi} \pm \dfrac{\sqrt{2\pi(A + 72\pi)}}{2\pi} \\[5ex] r = \dfrac{-12\pi \pm \sqrt{2\pi(A + 72\pi)}}{2\pi} \\[7ex] (c.) \\[3ex] A = 300\;in^2 \\[3ex] r = \dfrac{-12\pi \pm \sqrt{144\pi^2 + 2\pi * 300}}{2\pi} \\[5ex] r = \dfrac{-12\pi \pm \sqrt{144\pi^2 + 600\pi}}{2\pi} \\[5ex] r = \dfrac{-37.69911184 \pm \sqrt{1421.223034 + 1884.955592}}{6.283185307} \\[5ex] r = \dfrac{-37.69911184 \pm \sqrt{3306.178626}}{6.283185307} \\[5ex] r = \dfrac{-37.69911184 \pm 57.49937935}{6.283185307} \\[5ex] r = \dfrac{-37.69911184 + 57.49937935}{6.283185307} \;\;OR\;\; r = \dfrac{-37.69911184 - 57.49937935}{6.283185307} \\[5ex] r = \dfrac{19.80026095}{6.283185307} \;\;OR\;\; r = \dfrac{-95.19849119}{6.283185307} \\[5ex] r = 3.151309405 \;\;OR\;\; -15.15130145 \\[3ex] r\;\;cannnot\;\;be\;\;negative \\[3ex] \therefore r = 3.151309405 \\[3ex] r \approx 3.15\;inches\;\;to\;\;two\;\;decimal\;\;places$
(18.) Indicate whether these set of two functions are equal.
If the two functions are not equal, specify the element of the domain for which the two functions have different values.

First Set:

$f: \mathbb{Z} \rightarrow \mathbb{Z}, \;\;where\;\; f(x) = x^4 \\[3ex] g: \mathbb{Z} \rightarrow \mathbb{Z}, \;\;where\;\; g(x) = |x|^4 \\[3ex]$ Second Set:

$f: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}, \;\;where\;\; f(x, y) = |x + y|^3 \\[3ex] g: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}, \;\;where\;\; g(x, y) = |x|^3 + |y|^3 \\[3ex]$

$\underline{1st\;\;set} \\[3ex] f: \mathbb{Z} \rightarrow \mathbb{Z}, \;\;where\;\; f(x) = x^4 \\[3ex] g: \mathbb{Z} \rightarrow \mathbb{Z}, \;\;where\;\; g(x) = |x|^4 \\[3ex] \underline{Test\;\;numbers:\;\;negative,zero,positive} \\[3ex] Let\;\; x = -1 \\[3ex] f(-1) = (-1)^4 = (-1)(-1)(-1)(-1) = 1 \\[3ex] g(-1) = |-1|^4 = 1^4 = 1 \\[3ex] f(-1) = g(-1) \\[3ex] Let\;\; x = 0 \\[3ex] f(0) = 0^4 = 0 \\[3ex] g(0) = |0|^4 = 0^4 = 0 \\[3ex] f(0) = g(0) \\[3ex] Let\;\; x = 2 \\[3ex] f(2) = 2^4 = 16 \\[3ex] g(2) = |-2|^4 = 2^4 = 16 \\[3ex] f(2) = g(2) \\[3ex] \therefore f(x) = g(x) \\[5ex] \underline{2nd\;\;set} \\[3ex] f: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}, \;\;where\;\; f(x, y) = |x + y|^3 \\[3ex] g: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}, \;\;where\;\; g(x, y) = |x|^3 + |y|^3 \\[3ex] \underline{Test\;\;numbers:\;\;negative,zero,positive} \\[3ex] Let\;\; x = -1, y = -2 \\[3ex] f(-1, -2) \\[3ex] = |-1 + -2|^3 \\[3ex] = |-1 - 2|^3 \\[3ex] = |-3|^3 \\[3ex] = 3^3 \\[3ex] = 27 \\[3ex] g(-1, -2) \\[3ex] = |-1|^3 + |-2|^3 \\[3ex] = 1^3 + 2^3 \\[3ex] = 1 + 8 \\[3ex] = 9 \\[3ex] 27 \ne 9 \\[3ex] f(-1, -2) \ne g(-1, -2) \\[3ex] \therefore f(x) \ne g(x) \\[3ex]$ You may decide to stop here.
(19.) The domain and target set of functions $f$ and $g$ is $\mathbb{R}$
The functions are defined as:

$f(x) = 2x - 3 \\[3ex] g(x) = 5x - 7 \\[3ex]$ Determine:

$(I.)\;\; f \circ g \\[3ex] (II.)\;\; g \circ f \\[3ex] (III.)\;\; (f \circ g)^{-1} \\[3ex] (IV.)\;\; (g \circ f)^{-1} \\[3ex] (V.)\;\; f^{-1} \circ g^{-1} \\[3ex] (VI.)\;\; g^{-1} \circ f^{-1} \\[3ex]$ $(VII.)$ Are any of the above the same?

$f(x) = 2x - 3 \\[3ex] g(x) = 5x - 7 \\[3ex] (I.) \\[3ex] f \circ g \\[3ex] = f(g(x)) \\[3ex] = f(5x - 7) \\[3ex] = 2(5x - 7) - 3 \\[3ex] = 10x - 14 - 3 \\[3ex] = 10x - 17 \\[3ex] (II.) \\[3ex] g \circ f \\[3ex] = g(f(x)) \\[3ex] = g(2x - 3) \\[3ex] = 5(2x - 3) - 7 \\[3ex] = 10x - 15 - 7 \\[3ex] = 10x - 22 \\[3ex] (III.) \\[3ex] (f \circ g)^{-1} \\[3ex] = (10x - 17)^{-1} \\[3ex] Let\;\; y = 10x - 17 \\[3ex] Interchange\;\;x\;\;and\;\;y \\[3ex] x = 10y - 17 \\[3ex] Solve\;\;for\;\;y \\[3ex] 10y - 17 = x \\[3ex] 10y = x + 17 \\[3ex] y = \dfrac{x + 17}{10} \\[5ex] (f \circ g)^{-1} = \dfrac{x + 17}{10} \\[5ex] (IV.) \\[3ex] (g \circ f)^{-1} \\[3ex] = (10x - 22)^{-1} \\[3ex] Let\;\; y = 10x - 22 \\[3ex] Interchange\;\;x\;\;and\;\;y \\[3ex] x = 10y - 22 \\[3ex] Solve\;\;for\;\;y \\[3ex] 10y - 22 = x \\[3ex] 10y = x + 22 \\[3ex] y = \dfrac{x + 22}{10} \\[5ex] (g \circ f)^{-1} = \dfrac{x + 22}{10} \\[5ex] f(x) = 2x - 3 \\[3ex] f^{-1}(x) = (2x - 3)^{-1} \\[3ex] \\[3ex] Let\;\; y = 2x - 3 \\[3ex] Interchange\;\;x\;\;and\;\;y \\[3ex] x = 2y - 3 \\[3ex] Solve\;\;for\;\;y \\[3ex] 2y - 3 = x \\[3ex] 2y = x + 3 \\[3ex] y = \dfrac{x + 3}{2} \\[5ex] f^{-1}(x) = \dfrac{x + 3}{2} \\[5ex] g(x) = 5x - 7 \\[3ex] g^{-1}(x) = (5x - 7)^{-1} \\[3ex] \\[3ex] Let\;\; y = 5x - 7 \\[3ex] Interchange\;\;x\;\;and\;\;y \\[3ex] x = 5y - 7 \\[3ex] Solve\;\;for\;\;y \\[3ex] 5y - 7 = x \\[3ex] 5y = x + 7 \\[3ex] y = \dfrac{x + 7}{5} \\[5ex] g^{-1}(x) = \dfrac{x + 7}{5} \\[5ex] (V.) \\[3ex] f^{-1}(x) = \dfrac{x + 3}{2} \\[5ex] g^{-1}(x) = \dfrac{x + 7}{5} \\[5ex] f^{-1} \circ g^{-1} \\[3ex] = f^{-1}(g^{-1}) \\[3ex] = f^{-1}\left(\dfrac{x + 7}{5}\right) \\[5ex] = \dfrac{\left(\dfrac{x + 7}{5}\right) + 3}{2} \\[7ex] = \left(\dfrac{x + 7}{5}\right) + \dfrac{3}{1} \div \dfrac{2}{1} \\[5ex] = \dfrac{x + 7 + 15}{5} * \dfrac{1}{2} \\[5ex] f^{-1} \circ g^{-1} = \dfrac{x + 22}{10} \\[5ex] (VI.) \\[3ex] f^{-1}(x) = \dfrac{x + 3}{2} \\[5ex] g^{-1}(x) = \dfrac{x + 7}{5} \\[5ex] g^{-1} \circ f^{-1} \\[3ex] = g^{-1}(f^{-1}) \\[3ex] = g^{-1}\left(\dfrac{x + 3}{2}\right) \\[5ex] = \dfrac{\left(\dfrac{x + 3}{2}\right) + 7}{5} \\[7ex] = \left(\dfrac{x + 3}{2}\right) + \dfrac{7}{1} \div \dfrac{5}{1} \\[5ex] = \dfrac{x + 3 + 14}{2} * \dfrac{1}{5} \\[5ex] = \dfrac{x + 17}{10} \\[5ex] g^{-1} \circ f^{-1} = \dfrac{x + 17}{10} \\[5ex] (VII.) \\[3ex] f^{-1} \circ g^{-1} = (g \circ f)^{-1} = \dfrac{x + 22}{10} \\[5ex] g^{-1} \circ f^{-1} = (f \circ g)^{-1} = \dfrac{x + 17}{10}$
(20.) (a.) Given the function: f(x) = 3x + 2, determine the rate of change between 4 and 7

(b.) Determine the rate of change between 2 and 5 for $f(x) = \sqrt{(2x - 3) + 5}$

(c.) Determine the rate of change between 2 and 5 for $f(x) = \sqrt{2x - 3} + 5$

(c.) Find the rate for change between 0 and 1 for the graph:

$rate\;\;of\;\;change = slope = m \\[3ex] (a.) \\[3ex] f(x) = 3x + 2 \\[3ex] x_1 = 4 \\[3ex] f(x_1) = 3(4) + 2 \\[3ex] f(x_1) = 12 + 2 \\[3ex] f(x_1) = 14 \\[3ex] x_2 = 7 \\[3ex] f(x_2) = 3(7) + 2 \\[3ex] f(x_2) = 21 + 2 \\[3ex] f(x_2) = 23 \\[3ex] m = \dfrac{f(x_2) - f(x_1)}{x_2 - x_1} \\[5ex] m = \dfrac{23 - 14}{7 - 4} \\[5ex] m = \dfrac{9}{3} \\[5ex] m = 3 \\[5ex] (b.) \\[3ex] f(x) = \sqrt{(2x - 3) + 5} \\[3ex] f(x) = \sqrt{2x - 3 + 5} \\[3ex] f(x) = \sqrt{2x + 2} \\[3ex] x_1 = 2 \\[3ex] f(x_1) = \sqrt{2(2) + 2} \\[3ex] f(x_1) = \sqrt{4 + 2} \\[3ex] f(x_1) = \sqrt{6} \\[3ex] x_2 = 5 \\[3ex] f(x_2) = \sqrt{2(5) + 2} \\[3ex] f(x_2) = \sqrt{10 + 2} \\[3ex] f(x_2) = \sqrt{12} \\[3ex] f(x_2) = \sqrt{4 * 3} \\[3ex] f(x_2) = \sqrt{4} * \sqrt{3} \\[3ex] f(x_2) = 2\sqrt{3} \\[3ex] m = \dfrac{f(x_2) - f(x_1)}{x_2 - x_1} \\[5ex] m = \dfrac{2\sqrt{3} - \sqrt{6}}{5 - 2} \\[5ex] m = \dfrac{2\sqrt{3} - \sqrt{6}}{3} \\[5ex] (c.) \\[3ex] f(x) = \sqrt{2x - 3} + 5 \\[3ex] x_1 = 2 \\[3ex] f(x_1) = \sqrt{2(2) - 3} + 5 \\[3ex] f(x_1) = \sqrt{4 - 3} + 5 \\[3ex] f(x_1) = \sqrt{1} + 5 \\[3ex] f(x_1) = 1 + 5 \\[3ex] f(x_1) = 6 \\[3ex] x_2 = 5 \\[3ex] f(x_2) = \sqrt{2(5) - 3} + 5 \\[3ex] f(x_2) = \sqrt{10 - 3} + 5 \\[3ex] f(x_2) = \sqrt{7} + 5 \\[3ex] f(x_2) = 5 + \sqrt{7} \\[3ex] m = \dfrac{f(x_2) - f(x_1)}{x_2 - x_1} \\[5ex] m = \dfrac{(5 + \sqrt{7}) - 6}{5 - 2} \\[5ex] m = \dfrac{5 + \sqrt{7} - 6}{3} \\[5ex] m = \dfrac{-1 + \sqrt{7}}{3} \\[5ex] m = \dfrac{\sqrt{7} - 1}{3} \\[5ex] (d.) \\[3ex] x_1 = 0 \\[3ex] y_1 = 0 \\[3ex] x_2 = 1 \\[3ex] y_2 = \dfrac{4 + 8}{2} \\[5ex] y_2 = \dfrac{12}{2} \\[5ex] y_2 = 6 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{6 - 0}{1 - 0} \\[5ex] m = \dfrac{6}{1} \\[5ex] m = 6$

(21.)

(22.)

(23.)

(24.)

(25.)

(26.)

(27.)

(28.)

(29.)

(30.)

(31.)

(32.)

(33.)

(34.)