Samuel Dominic Chukwuemeka (SamDom For Peace)

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka



Solved Examples on the Evaluation of Functions

Pre-requisites:
Linear Equations
Literal Equations
All previous topics in Relations and 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 any wrong answer.

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

Solve all questions.
Show all work.

(1.) CSEC The functions $f(x)$ and $g(x)$ are defined as:

$ f(x) = \dfrac{5x - 4}{3} \:\:\:\:\: g(x) = x^2 - 1 \\[5ex] $ (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 $fg(x)$


$ (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} $
(2.) 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} $
(3.) 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 $
(4.) 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 $
(5.) CSEC The functions $f(x)$ and $g(x)$ are defined as

$ f(x) = 3x + 2 \:\:\:\:\:\:\:\:\:\:\: 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) $
(6.) $f(x) = \sqrt{5x + 3}$

Calculate $f\left(\dfrac{13}{5}\right),\:\:\: f(12),\:\:\: f\left(-\dfrac{2}{5}\right)$


$f\left(\dfrac{13}{5}\right)$
This means that we have to substitute $\dfrac{13}{5}$ for $x$

$ f(x) = \sqrt{5x + 3} \\[3ex] f\left(\dfrac{13}{5}\right) = \sqrt{5 * {\dfrac{13}{5}} + 3} \\[3ex] = \sqrt{13 + 3} \\[3ex] = \sqrt{16} \\[3ex] = 4 $

$f(12)$
This means that we have to substitute $12$ for $x$

$ f(x) = \sqrt{5x + 3} \\[3ex] f(12) = \sqrt{5 * 12 + 3} \\[3ex] = \sqrt{60 + 3} \\[3ex] = \sqrt{63} \\[3ex] = \sqrt{9 * 7} \\[3ex] = \sqrt{9} * \sqrt{7} \\[3ex] = 3 * \sqrt{7} \\[3ex] = 3\sqrt{7} $

$f\left(-\dfrac{2}{5}\right)$
This means that we have to substitute $-\dfrac{2}{5}$ for $x$

$ f(x) = \sqrt{5x + 3} \\[3ex] f\left(-\dfrac{2}{5}\right) = \sqrt{5 * -\dfrac{2}{5} + 3} \\[3ex] = \sqrt{-2 + 3} \\[3ex] = \sqrt{1} \\[3ex] = 1 $
(7.) $g(x) = \dfrac{x}{\sqrt{1 - x^2}}$

Calculate $g(0),\:\:\: g(-1),\:\:\: g(5),\:\:\: g\left(\dfrac{2}{3}\right)$


$g(0)$
This means that we have to substitute $0$ for $x$

$ g(x) = \dfrac{x}{\sqrt{1 - x^2}} \\[3ex] g(0) = \dfrac{0}{\sqrt{1 - 0^2}} \\[3ex] g(0) = \dfrac{0}{\sqrt{1 - 0}} \\[3ex] g(0) = \dfrac{0}{\sqrt{1}} \\[3ex] g(0) = \dfrac{0}{1} \\[3ex] g(0) = 0 $

$g(-1)$
This means that we have to substitute $-1$ for $x$

$ g(x) = \dfrac{x}{\sqrt{1 - x^2}} \\[3ex] g(-1) = \dfrac{-1}{\sqrt{1 - (-1)^2}} \\[3ex] (-1)^2 = -1 * -1 = 1 \\[3ex] g(-1) = \dfrac{-1}{\sqrt{1 - 1}} \\[3ex] g(-1) = \dfrac{-1}{\sqrt{0}} \\[3ex] g(-1) = \dfrac{-1}{0} \\[3ex] g(-1) = undefined $
Can you divide "anything" by "nothing"?
Can you create "anything" out of nothing?
Only GOD can do that!

$g(5)$
This means that we have to substitute $5$ for $x$

$ g(x) = \dfrac{x}{\sqrt{1 - x^2}} \\[3ex] g(5) = \dfrac{5}{\sqrt{1 - 5^2}} \\[3ex] g(5) = \dfrac{5}{\sqrt{1 - 25}} \\[3ex] g(5) = \dfrac{5}{\sqrt{-24}} $
$g(5)$ is not a real number.
It is an imaginary number.
All the numbers we shall cover in this topic are only real numbers.

$g\left(\dfrac{2}{3}\right)$
This means that we have to substitute $\dfrac{2}{3}$ for $x$

$ g(x) = \dfrac{x}{\sqrt{1 - x^2}} \\[5ex] g\left(\dfrac{2}{3}\right) = \dfrac{\dfrac{2}{3}}{\sqrt{1 - \left(\dfrac{2}{3}\right)^2}} \\[9ex] g\left(\dfrac{2}{3}\right) = \dfrac{\dfrac{2}{3}}{\sqrt{1 - \left(\dfrac{2^2}{3^2}\right)}} \\[9ex] g\left(\dfrac{2}{3}\right) = \dfrac{\dfrac{2}{3}}{\sqrt{1 - \dfrac{4}{9}}} \\[9ex] g\left(\dfrac{2}{3}\right) = \dfrac{\dfrac{2}{3}}{\sqrt{\dfrac{9}{9} - \dfrac{4}{9}}} \\[9ex] g\left(\dfrac{2}{3}\right) = \dfrac{\dfrac{2}{3}}{\sqrt{\dfrac{9 - 4}{9}}} \\[9ex] g\left(\dfrac{2}{3}\right) = \dfrac{\dfrac{2}{3}}{\sqrt{\dfrac{5}{9}}} \\[9ex] g\left(\dfrac{2}{3}\right) = \dfrac{2}{3} \div \sqrt{\dfrac{5}{9}} \\[7ex] g\left(\dfrac{2}{3}\right) = \dfrac{2}{3} \div \dfrac{\sqrt{5}}{\sqrt{9}} \\[7ex] g\left(\dfrac{2}{3}\right) = \dfrac{2}{3} \div \dfrac{\sqrt{5}}{3} \\[7ex] g\left(\dfrac{2}{3}\right) = \dfrac{2}{3} * \dfrac{3}{\sqrt{5}} \\[7ex] g\left(\dfrac{2}{3}\right) = \dfrac{2}{\sqrt{5}} \\[7ex] g\left(\dfrac{2}{3}\right) = \dfrac{2}{\sqrt{5}} * \dfrac{\sqrt{5}}{\sqrt{5}} \\[7ex] g\left(\dfrac{2}{3}\right) = \dfrac{2\sqrt{5}}{5} $
(8.) $f(x) = -4x + 7$

Calculate $f(a),\:\:\: f(a + 2),\:\:\: f(a + h)$


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

$ f(x) = -4x + 7 \\[3ex] f(a) = -4a + 7 $

$f(a + 2)$
This means that we have to substitute $(a + 2)$ for $x$

$ f(x) = -4x + 7 \\[3ex] f(a + 2) = -4(a + 2) + 7 \\[3ex] = -4a - 8 + 7 \\[3ex] = -4a - 1 $

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

$ f(x) = -4x + 7 \\[3ex] f(a + h) = -4(a + h) + 7 \\[3ex] = -4a - 4h + 7 $
(9.) $g(x) = \dfrac{x - 5}{x + 2}$

Calculate $g(6),\:\:\: g(5),\:\:\: g(-2),\:\:\: g(-12.75),\:\:\: g(x + h)$


$g(6)$
This means that we have to substitute $6$ for $x$

$ g(x) = \dfrac{x - 5}{x + 2} \\[5ex] g(6) = \dfrac{6 - 5}{6 + 2} \\[5ex] = \dfrac{1}{8} $

$g(5)$
This means that we have to substitute $5$ for $x$

$ g(x) = \dfrac{x - 5}{x + 2} \\[5ex] g(5) = \dfrac{5 - 5}{5 + 2} \\[5ex] = \dfrac{0}{7} \\[3ex] = 0 $

$g(-2)$
This means that we have to substitute $(-2)$ for $x$

$ g(x) = \dfrac{x - 5}{x + 2} \\[5ex] g(-2) = \dfrac{-2 - 5}{-2 + 2} \\[5ex] = -\dfrac{7}{0} \\[5ex] = undefined \\[3ex] g(-2) DNE $

$g(-12.75)$
This means that we have to substitute $(-12.75)$ for $x$

$ g(x) = \dfrac{x - 5}{x + 2} \\[5ex] g(-12.75) = \dfrac{-12.75 - 5}{-12.75 + 2} \\[5ex] = \dfrac{-17.75}{-10.75} \\[5ex] = 1.651162791 $

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

$ g(x) = \dfrac{x - 5}{x + 2} \\[5ex] g(x + h) = \dfrac{x + h - 5}{x + h + 2} \\[5ex] = \dfrac{x + h - 5}{x + h + 2} $
(10.) $f(x) = x^2 - 3x$

Calculate $f(-a),\:\:\: f(a - 3),\:\:\: f(a + h)$


$f(-a)$
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.) If $f(x) = 4x^3 + Ax^2 + 7x - 5$ and $f(2) = -7$
Calculate the value of $A$


$ f(x) = 4x^3 + Ax^2 + 7x - 5 \\[3ex] f(2) = 4(2)^3 + A(2)^2 + 7(2) - 5 \\[3ex] f(2) = 4(8) + A(4) + 14 - 5 \\[3ex] f(2) = 32 + 4A + 9 \\[3ex] f(2) = 4A + 41 \\[5ex] But\:\: f(2) = -7 \\[3ex] -7 = 4A + 41 \\[3ex] 4A + 41 = -7 \\[3ex] 4A = -7 - 41 \\[3ex] 4A = -48 \\[3ex] A = -\dfrac{48}{4} \\[3ex] A = -12 $
(12.) $f(x) = -x^2 - 3x - 12$
Calculate $f(-a),\:\:\: f(-a - 2),\:\:\: f(a + 7)$


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

$ f(x) = -x^2 - 3x - 12 \\[2ex] f(x) = -1 * x^2 - 3x - 12 \\[2ex] f(-a) = -1 * (-a)^2 - 3(-a) - 12 \\[2ex] f(-a) = -1 * a^2 + 3a - 12 \\[2ex] f(-a) = -a^2 + 3a - 12 \\[2ex] f(-a) = -1(a^2 - 3a + 12) $

$f(-a - 2)$
This means that we have to substitute $-a - 2$ for $x$

$ f(x) = -x^2 - 3x - 12 \\[3ex] f(x) = -1 * x^2 - 3x - 12 \\[3ex] f(-a - 2) = -1 * (-a - 2)^2 - 3(-a - 2) - 12 \\[3ex] (-a - 2)^2 = (-a - 2)(-a - 2) \\[3ex] = a^2 + 2a + 2a + 4 \\[3ex] = a^2 + 4a + 4 \\[3ex] f(-a - 2) = -1(a^2 + 4a + 4) - 3(-a - 2) - 12 \\[3ex] f(-a - 2) = -a^2 - 4a - 4 + 3a + 6 - 12 \\[3ex] f(-a - 2) = -a^2 - a - 10 \\[3ex] f(-a - 2) = -1(a^2 + a + 10) $

$f(a + 7)$
This means that we have to substitute $a + 7$ for $x$

$ f(x) = -x^2 - 3x - 12 \\[3ex] f(x) = -1 * x^2 - 3x - 12 \\[3ex] f(a + 7) = -1 * (a + 7)^2 - 3(a + 7) - 12 \\[3ex] (a + 7)^2 = (a + 7)(a + 7) \\[3ex] = a^2 + 7a + 7a + 49 \\[3ex] = a^2 + 14a + 49 \\[3ex] f(a + 7) = -1(a^2 + 14a + 49) - 3(a + 7) - 12 \\[3ex] f(a + 7) = -a^2 - 14a - 49 - 3a - 21 - 12 \\[3ex] f(a + 7) = -a^2 - 17a - 82 \\[3ex] f(a + 7) = -1(a^2 + 17a + 82) $
(13.) ACT A function, $f$ is defined by $f(x, y) = 3x^2 - 4y$
What is the value of $f(4, 3)$?


$f(4, 3)$
This means that we have to substitute $4$ for $x$ and $3$ for $y$

$ f(x, y) = 3x^2 - 4y \\[3ex] f(4, 3) = 3(4)^2 - 4(3) \\[3ex] f(4, 3) = 3(16) - 12 \\[3ex] f(4, 3) = 48 - 12 \\[3ex] f(4,3) = 36 $
(14.) ACT Let the function $f$ be defined as $f(x) = 5x^2 - 7(4x + 3)$.
What is the value of $f(3)$?


$f(3)$
This means that we have to substitute $4$ for $x$ and $3$ for $y$

$ f(x) = 5x^2 - 7(4x + 3) \\[3ex] f(x) = 5 * x^2 - 7(4 * x + 3) \\[3ex] f(3) = 5 * 3^2 - 7(4 * 3 + 3) \\[3ex] f(3) = 5 * 9 - 7(12 + 3) \\[3ex] f(3) = 45 - 7(15) \\[3ex] f(3) = 45 - 105 \\[3ex] f(3) = -60 $
(15.) ACT The function $f(x)$ is shown below with several points labeled.
Another function, $g(x)$ is defined such that $g(x) = -[f(x) - 3]$.
What is $g(4)$?
evaluating functions - Q15
$ F.\:\: -4 \\[3ex] G.\:\: -1 \\[3ex] H.\:\: 1 \\[3ex] J.\:\: 4 \\[3ex] K.\:\: 7 \\[3ex] $

$ g(x) = -[f(x) - 3] \\[3ex] For\:\:g(4),\:\:x = 4 \\[3ex] g(4) = -[f(4) - 3] \\[3ex] f(4) = -1 \\[3ex] \rightarrow g(4) = -[-1 - 3] \\[3ex] g(4) = -(-4) \\[3ex] g(4) = 4 $
(16.) ACT Let a function of $2$ variables be defined by $f(x, y) = xy - (x - y)$.
What is the value of $f(10, 3)$?


$f(10, 3)$
This means that we have to substitute $10$ for $x$ and $3$ for $y$

$ f(x, y) = xy - (x - y) \\[3ex] f(10, 3) = 10(3) - (10 - 3) \\[3ex] f(10, 3) = 30 - 7 \\[3ex] f(10, 3) = 23 $