Solved Examples on the Composition of Functions

First Method (long way)
$ (f \circ g)(x) \\[3ex] = f(g(x)) \\[3ex] = f(x^2 - 1) \\[3ex] But, f(x) = -3x - 1 \\[3ex] f(x^2 - 1) = -3(x^2 -1) - 1 \\[3ex] = -3x^2 + 3 - 1 \\[3ex] = -3x^2 + 2 \\[3ex] \therefore (f \circ g)(x) = -3x^2 + 2 \\[3ex] (f \circ g)(1) = -3(1)^2 + 2 \\[3ex] = -3(1) + 2 \\[3ex] = -3 + 2 \\[3ex] = -1 \\[3ex] (f \circ g)(1) = -1 \\[3ex] $ Second Method (short way)
$ (f \circ g)(1) \\[3ex] = f(g(1)) \\[3ex] g(x) = x^2 - 1 \\[3ex] g(1) = 1^2 - 1 \\[3ex] g(1) = 1 - 1 \\[3ex] g(1) = 0 \\[3ex] $ We now need to find $f(0)$
$ f(x) = -3x - 1 \\[3ex] f(0) = -3(0) - 1 \\[3ex] f(0) = 0 - 1 \\[3ex] f(0) = -1 \\[3ex] (f \circ g)(1) = -1 $

$ (f \circ g)(x) \\[3ex] = f(g(x)) \\[3ex] = f(7x - 3) \\[3ex] But, f(x) = 5x \\[3ex] f(7x - 3) = 5(7x - 3) \\[3ex] = 35x - 15 \\[3ex] (f \circ g)(x) = 35x - 15 \\[3ex] $ The domain of $(f \circ g)(x)$ is all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$
Let us look at $g(x)$
$ g(x) = 7x - 3 \\[3ex] D = (-\infty, \infty) \\[3ex] $ Next, we look at $(f \circ g)(x)$
$ (f \circ g)(x) = 35x - 15 \\[3ex] D = (-\infty, \infty) \\[3ex] $ $\therefore D = (-\infty, \infty)$

$ (g \circ f)(x) \\[3ex] = g(f(x)) \\[3ex] = g(5x) \\[3ex] But, g(x) = 7x - 3 \\[3ex] g(5x) = 7(5x) - 3 \\[3ex] = 35x - 3 \\[3ex] (g \circ f)(x) = 35x - 3 \\[3ex] $ The domain of $(g \circ f)(x)$ is all $x$ in the domain of $f$ such that $f(x)$ is in the domain of $g$
Let us look at $f(x)$
$f(x) = 5x$
There is no restriction in this domain.
Any input value will give an output.
$D = (-\infty, \infty)$

Next, we look at $(g \circ f)(x)$
$(g \circ f)(x) = 35x - 3$
Each input value in this domain will give an output value.
$D = (-\infty, \infty)$

$\therefore D = (-\infty, \infty)$
$D$ = {$x | x \in \mathbb{R}$}

First Method (long way)
$ (g \circ g)(x) \\[3ex] = g(g(x)) \\[3ex] = g(x^2 - 4x - 9) \\[3ex] But, g(x) = x^2 - 4x - 9 \\[3ex] g(x^2 - 4x - 9) = (x^2 - 4x - 9)^2 - 4(x^2 - 4x - 9) - 9 \\[3ex] = (x^2 - 4x - 9)(x^2 - 4x - 9) - 4x^2 + 16x + 36 - 9 \\[3ex] = x^4 - 4x^3 - 9x^2 - 4x^3 + 16x^2 + 36x - 9x^2 + 36x + 81 - 4x^2 + 16x + 27 \\[3ex] = x^4 - 8x^3 - 6x^2 + 88x + 108 \\[3ex] \therefore (g \circ g)(x) = x^4 - 8x^3 - 6x^2 + 88x + 108 \\[3ex] (g \circ g)(-3) = (-3)^4 - 8 * (-3)^3 - 6 * (-3)^2 + 88 * (-3) + 108 \\[3ex] = 81 - 8(-27) - 6(9) - 264 + 108 \\[3ex] = 81 + 216 - 54 - 264 + 108 \\[3ex] = 87 \\[3ex] (g \circ g)(-3) = 87 \\[3ex] $ Second Method (short way)
$ (g \circ g)(-3) \\[3ex] = g(g(-3)) \\[3ex] But, g(x) = x^2 - 4x - 9 \\[3ex] g(-3) = (-3)^2 - 4(-3) - 9 \\[3ex] = 9 + 12 - 9 \\[3ex] = 12 \\[3ex] $ We now need to find $g(12)$
$ g(x) = x^2 - 4x - 9 \\[3ex] g(12) = (12)^2 - 4(12) - 9 \\[3ex] = 144 - 48 - 9 \\[3ex] = 87 \\[3ex] (g \circ g)(-3) = 87 \\[3ex] $

First Method (long way)
$ (f \circ f)(x) \\[3ex] = f(f(x)) \\[3ex] = f(x^3) \\[3ex] But, f(x) = x^3 \\[3ex] f(x^3) = (x^3)^3 \\[3ex] = x^9 \\[3ex] \therefore (f \circ f)(x) = x^9 \\[3ex] (f \circ f)(-2) = (-2)^9 \\[3ex] = -512 \\[3ex] (f \circ f)(-2) = -512 \\[3ex] $ Second Method (short way)
$ (f \circ f)(-2) \\[3ex] = f(f(-2)) \\[3ex] But, f(x) = x^3 \\[3ex] f(-2) = (-2)^3 \\[3ex] = -8 \\[3ex] $ We now need to find $f(-8)$
$ f(x) = x^3 \\[3ex] f(-8) = (-8)^3 \\[3ex] = -512 \\[3ex] (f \circ f)(-2) = -512 $

$ (f \circ g)(x) \\[3ex] = f(g(x)) \\[3ex] = f\left(\dfrac{1}{x}\right) \\[5ex] But, f(x) = \dfrac{3}{1 - 7x} \\[5ex] f\left(\dfrac{1}{x}\right) = \dfrac{3}{1 - 7\left(\dfrac{1}{x}\right)} \\[7ex] = \dfrac{3}{1 - \dfrac{7}{x}} \\[7ex] = \dfrac{3}{\dfrac{x}{x} - \dfrac{7}{x}} \\[7ex] = \dfrac{3}{\dfrac{x - 7}{x}} \\[7ex] = 3 \div \dfrac{x - 7}{x} \\[7ex] = 3 * \dfrac{x}{x - 7} \\[7ex] = \dfrac{3x}{x - 7} \\[5ex] (f \circ g)(x) = \dfrac{3x}{x - 7} $

The domain of $(f \circ g)(x)$ is all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$
Let us look at $g(x)$
$g(x) = \dfrac{1}{x}$

$x$ cannot be equal to $0$
This is because that would make the function to be undefined.
You cannot divide anything by $0$
This means that $0$ is excluded from the domain.
$0$ is not in the domain of $g(x)$
Hence, $0$ cannot be in the domain of f(g(x))

Next, we look at $(f \circ g)(x) = \dfrac{3x}{x - 7} \\[3ex]$ $x - 7$ cannot be $0$
This is because that would make the function to be undefined.
You cannot divide anything by $0$
$x$ cannot be equal to $7$
$7$ is not in the domain of f(g(x))
$\therefore D = (-\infty, 0) \cup (0, 7) \cup (7, \infty)$



$ (g \circ f)(x) \\[3ex] = g(f(x)) \\[3ex] = g\left(\dfrac{3}{1 - 7x}\right) \\[5ex] But, g(x) = \dfrac{1}{x} \\[5ex] g\left(\dfrac{3}{1 - 7x}\right) = \dfrac{1}{\dfrac{3}{1- 7x}} \\[7ex] = 1 \div \dfrac{3}{1 - 7x} \\[7ex] = 1 * \dfrac{1- 7x}{3} \\[7ex] = \dfrac{1- 7x}{3} \\[5ex] (g \circ f)(x) = \dfrac{1- 7x}{3} $

The domain of $(g \circ f)(x)$ is all $x$ in the domain of $f$ such that $f(x)$ is in the domain of $g$
Let us look at $f(x)$
$f(x) = \dfrac{3}{1 - 7x} \\$
$1 - 7x$ cannot be equal to $0$
This is because that would make the function to be undefined.
You cannot divide anything by $0$
Just set it to be equal to zero and solve.
$ 1 - 7x = 0 \\[3ex] 1 = 0 + 7x \\[3ex] 1 = 7x \\[3ex] 7x = 1 \\[3ex] x = \dfrac{1}{7} \\[5ex] $ So, $x$ cannot be equal to $\dfrac{1}{7}$

This means that $\dfrac{1}{7}$ is excluded from the domain.

$\dfrac{1}{7}$ is not in the domain of $f(x)$

Hence, $\dfrac{1}{7}$ cannot be in the domain of g(f(x))

Next, we look at $(g \circ f)(x) = \dfrac{1- 7x}{3}$
There is no restriction in the domain of $(g \circ f)(x)$
Any real number can be in the domain of $g(f(x))$
$\therefore D = \left(-\infty, \dfrac{1}{7}\right) \cup \left(\dfrac{1}{7}, \infty\right)$

$ (f \circ g)(x) \\[3ex] = f(g(x)) \\[3ex] = f(3x - 4) \\[3ex] But, f(x) = \sqrt{x + 5} \\[3ex] f(3x - 4) = \sqrt{(3x - 4) + 5} \\[3ex] = \sqrt{3x - 4 + 5} \\[3ex] = \sqrt{3x + 1} \\[3ex] (f \circ g)(x) = \sqrt{3x + 1} \\[3ex] $ The domain of $(f \circ g)(x)$ is all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$
Let us look at $g(x)$
$g(x) = 3x - 4$
There is no restriction in this domain.
All real number input values will give output values.
$D = (-\infty, \infty)$

Next, we look at $(f \circ g)(x)$
$(f \circ g)(x) = \sqrt{3x + 1}$
$3x + 1$ cannot be a negative number
This is because the square root of any negative number is an imaginary number (not a real number).
We are only concerned with real numbers.
It has to be "at least" a nonnegative number.
Let us set the radicand (the term inside the radical) to be "at least" a nonnegative number
Nonnegative numbers includes zero and the positive numbers.
$ 3x + 1 \ge 0 \\[3ex] 3x \ge 0 - 1 \\[3ex] 3x \ge -1 \\[3ex] x \ge -\dfrac{1}{3} \\[5ex] D = \left[-\dfrac{1}{3}, \infty\right) \\[5ex] $ $\therefore D = \left[-\dfrac{1}{3}, \infty\right)$

$ (g \circ f)(x) \\[3ex] = g(f(x)) \\[3ex] = g(\sqrt{x + 5}) \\[3ex] But, g(x) = 3x - 4 \\[3ex] g(\sqrt{x + 5}) = 3(\sqrt{x + 5}) - 4 \\[3ex] = 3\sqrt{x + 5} - 4 \\[3ex] (g \circ f)(x) = 3\sqrt{x + 5} - 4 \\[3ex] $ The domain of $(g \circ f)(x)$ is all $x$ in the domain of $f$ such that $f(x)$ is in the domain of $g$
Let us look at $f(x)$
$f(x) = \sqrt{x + 5}$
$x + 5$ cannot be a negative number
This is because the square root of any negative number is an imaginary number (not a real number).
We are only concerned with real numbers.
It has to be "at least" a nonnegative number.
Let us set the radicand (the term inside the radical) to be "at least" a nonnegative number
Nonnegative numbers includes zero and the positive numbers.
$ x + 5 \ge 0 \\[3ex] x \ge 0 - 5 \\[3ex] x \ge -5 \\[3ex] D = [-5, \infty) \\[3ex] $ Next, we look at $(g \circ f)(x)$
$(g \circ f)(x) = 3\sqrt{x + 5} - 4$
$x + 5$ cannot be a negative number
This is because the square root of any negative number is an imaginary number (not a real number).
We are only concerned with real numbers.
It has to be "at least" a nonnegative number.
Let us set the radicand (the term inside the radical) to be "at least" a nonnegative number
Nonnegative numbers includes zero and the positive numbers.
$ x + 5 \ge 0 \\[3ex] x \ge 0 - 5 \\[3ex] x \ge -5 \\[3ex] $ $D = [-5, \infty)$

$\therefore D = [-5, \infty)$
$D$ = {$x | x \ge 5$}

$ (f \circ g)(x) \\[3ex] = f(g(x)) \\[3ex] = f(x^2 - 4) \\[3ex] But, f(x) = \sqrt{x + 4} \\[3ex] f(x^2 - 4) = \sqrt{(x^2 - 4) + 4} \\[3ex] = \sqrt{x^2 - 4 + 4} \\[3ex] = \sqrt{x^2} \\[3ex] = |x| \\[3ex] (f \circ g)(x) = |x|\\[3ex] $ Student: Why is the answer the absolute value of $x$? Is it okay to write it as just $x$?
Teacher: I would accept $x$. However, we are interested in only the positive value of $x$
The absolute value of $x$ gives only the positive value of $x$ (because of the $\sqrt{x^2}$).


The domain of $(f \circ g)(x)$ is all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$
Let us look at $g(x)$
$g(x) = x^2 - 4$
There is no restriction in this domain.
Any real number input value will give a result.
$D = (-\infty, \infty)$

Next, we look at $(f \circ g)(x)$
$(f \circ g)(x) = x$
There is also no restriction in this domain.
Any real number input value will give an output value.
$D = (-\infty, \infty)$

$\therefore D = (-\infty, \infty)$

$ (g \circ f)(x) \\[3ex] = g(f(x)) \\[3ex] = g(\sqrt{x + 4}) \\[3ex] But, g(x) = x^2 - 4 \\[3ex] g(\sqrt{x + 4}) = (\sqrt{x + 4})^2 - 4 \\[3ex] = x + 4 - 4 \\[3ex] = x \\[3ex] (g \circ f)(x) = x \\[3ex] $ The domain of $(g \circ f)(x)$ is all $x$ in the domain of $f$ such that $f(x)$ is in the domain of $g$
Let us look at $f(x)$
$f(x) = \sqrt{x + 4}$
$x + 4$ cannot be a negative number
This is because the square root of any negative number is an imaginary number (not a real number).
We are only concerned with real numbers.
It has to be "at least" a nonnegative number.
Let us set the radicand (the term inside the radical) to be "at least" a nonnegative number
Nonnegative numbers includes zero and the positive numbers.
$ x + 4 \ge 0 \\[3ex] x \ge 0 - 4 \\[3ex] x \ge -4 \\[3ex] $ $D = [-4, \infty)$

Next, we look at $(g \circ f)(x)$
$(g \circ f)(x) = x$
Each input value in this domain will give an output value.
$D = (-\infty, \infty)$

$\therefore D = [-4, \infty)$
$D$ = {$x | x \ge -4$}

$ f(x) = 9x^2 + 1 \\[3ex] g(x) = 4x + a \\[3ex] (f \circ g)(x) = f(g(x)) \\[3ex] = f(4x + a) \\[3ex] But,\:\: f(x) = 9x^2 + 1 \\[3ex] = 9(4x + a)^2 + 1 \\[3ex] = 9[(4x + a)(4x +a)] + 1 \\[3ex] = 9(16x^2 + 4ax + 4ax + a^2) + 1 \\[3ex] = 9(16x^2 + 8ax + a^2) + 1 \\[3ex] = 144x^2 + 72ax + 9a^2 + 1 \\[3ex] $ $f \circ g$ crosses the $y-axis$ at $730$

$144x^2 + 8ax + 9a^2 + 1 = 730$

On the $y-axis$, $x-values = 0$

$ 144(0)^2 + 72a(0) + 9a^2 + 1 = 730 \\[3ex] 9a^2 + 1 = 730 \\[3ex] 9a^2 = 730 - 1 \\[3ex] 9a^2 = 729 \\[3ex] a^2 = \dfrac{729}{9} \\[5ex] a^2 = 81 \\[3ex] a = \pm \sqrt{81} \\[3ex] a = \pm 9 \\[3ex] a = 9 \:\:OR\:\: a = -9 $

$ g(1) = 2 \\[3ex] f(g(1)) \\[3ex] = f(2) \\[3ex] = -6 $

$ g(f(5)) = ? \\[3ex] f(5) = 3 \\[3ex] g(3) = -5 \\[3ex] \therefore g(f(5)) = -5 $

$ g(f(3)) \\[3ex] f(3) = 2 \\[3ex] g(f(3)) = g(2) \\[3ex] g(2) = -3 $

$ g(f(5)) = ? \\[3ex] f(5) = 3 \\[3ex] g(3) = -5 \\[3ex] \therefore g(f(5)) = -5 $

$ f(x) = x^2 + 3x \\[3ex] g(x) = x + 1 \\[3ex] f(g(x)) = f(x + 1) \\[3ex] f(x + 1) \implies x = x + 1 \\[3ex] f(x + 1) = (x + 1)^2 + 3(x + 1) \\[3ex] f(x + 1) = (x + 1)(x + 1) + 3x + 3 \\[3ex] f(x + 1) = x^2 + x + x + 1 + 3x + 3 \\[3ex] f(x + 1) = x^2 + 5x + 2 $

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

$ f(x) = 9x^2 + 1 \\[3ex] g(x) = 4x + a \\[3ex] (f \circ g)(x) = f(g(x)) \\[3ex] = f(4x + a) \\[3ex] But,\:\: f(x) = 9x^2 + 1 \\[3ex] = 9(4x + a)^2 + 1 \\[3ex] = 9[(4x + a)(4x +a)] + 1 \\[3ex] = 9(16x^2 + 4ax + 4ax + a^2) + 1 \\[3ex] = 9(16x^2 + 8ax + a^2) + 1 \\[3ex] = 144x^2 + 8ax + 9a^2 + 1 \\[3ex] $ $f \circ g$ crosses the $y-axis$ at $730$

$144x^2 + 8ax + 9a^2 + 1 = 730$

On the $y-axis$, $x-values = 0$

$ 144(0)^2 + 8a(0) + 9a^2 + 1 = 730 \\[3ex] 9a^2 + 1 = 730 \\[3ex] 9a^2 = 730 - 1 \\[3ex] 9a^2 = 729 \\[3ex] a^2 = \dfrac{729}{9} \\[5ex] a^2 = 81 \\[3ex] a = \pm \sqrt{81} \\[3ex] a = \pm 9 \\[3ex] a = 9 \:\:OR\:\: a = -9 $

$ f(x) = \sqrt{x} \\[3ex] g(x) = 10x + b \\[3ex] (f \circ g)(x) \\[3ex] = f(g(x)) \\[3ex] = f(10x + b) \\[3ex] But, f(x) = \sqrt{x} \\[3ex] f(10x + b) = \sqrt{10x + b} \\[3ex] y = (f \circ g)(x) = \sqrt{10x + b} \\[3ex] Passes\:\:through\:\:(4, 6) \\[3ex] x = 4 \\[3ex] y = 6 \\[3ex] \implies 6 = \sqrt{10(4) + b} \\[3ex] \sqrt{40 + b} = 6 \\[3ex] Square\:\:both\:\:sides \\[3ex] 40 + b = 6^2 \\[3ex] 40 + b = 36 \\[3ex] b = 36 - 40 \\[3ex] b = -4 $