# Solved Examples on the Inverses of Functions

For each question as applicable, unless otherwise stated:
(I.) Determine the inverse of these functions.
(II.) Test for inverses.

(1.) $f(x) = 7x - 3$

$f(x) = 7x - 3 \\[3ex] y = 7x - 3 \\[3ex] Interchange\:\: x\:\: and\:\: y \\[3ex] x = 7y - 3 \\[3ex] Solve\:\: for\:\: y \\[3ex] 7y - 3 = x \\[3ex] 7y = x + 3 \\[3ex] y = \dfrac{x + 3}{7} \\[5ex] \therefore f^{-1}(x) = \dfrac{x + 3}{7} \\[5ex]$ Test for Inverses

$(f \circ f^{-1})(x) \\[3ex] = f(f^{-1}(x)) \\[3ex] = f\left(\dfrac{x + 3}{7}\right) \\[5ex] But\:\: f(x) = 7x - 3 \\[5ex] So\:\: f\left(\dfrac{x + 3}{7}\right) = 7\left(\dfrac{x + 3}{7}\right) - 3 \\[5ex] = x + 3 - 3 \\[5ex] = x \\[7ex] (f^{-1} \circ f)(x) \\[3ex] = f^{-1}(f(x)) \\[3ex] = f^{-1}(7x - 3) \\[3ex] But\:\: f^{-1}(x) = \dfrac{x + 3}{7} \\[5ex] So\:\: f^{-1}(7x - 3) = \dfrac{(7x - 3) + 3}{7} \\[5ex] = \dfrac{7x - 3 + 3}{7} \\[5ex] = \dfrac{7x}{7} \\[5ex] = x \\[3ex] (f \circ f^{-1})(x) = (f^{-1} \circ f)(x) = x$
(2.) Use function composition to determine if $f(x)$ and $g(x)$ are inverse functions.

$f(x) = \sqrt[3]{x + 1} \\[3ex] g(x) = x^3 + 1 \\[3ex]$

Using function composition to determine if two functions are inverses is to Test for Inverses
This implies that $f(x)$ and $g(x)$ are inverses if $(f \circ g)(x) = (g \circ f)(x) = x$
The two conditions must be satisfied for them to be inverses of each other:
(1.) $(f \circ g)(x) = (g \circ f)(x)$: The composition of $f$ and $g$ with $x$ must be equal to the composition of $g$ and $f$ with $x$
(2.) $(f \circ g)(x) = (g \circ f)(x) = x$: That composition must be equal to $x$
If any of these conditions are not true, then both functions are not inverses of each other.

$f(x) = \sqrt[3]{x + 1} \\[3ex] g(x) = x^3 + 1 \\[3ex] (f \circ g)(x) \\[3ex] = f(g(x)) \\[3ex] = f(x^3 + 1) \\[3ex] = \sqrt[3]{x^3 + 1 + 1} \\[3ex] = \sqrt[3]{x^3 + 2} \\[3ex] \ne x \\[3ex]$ This already implies that $f(x)$ and $g(x)$ are not inverse functions.
However, let us go ahead and find the composition of $g$ and $f$ with $x$

$(g \circ f)(x) \\[3ex] = g(f(x)) \\[3ex] = g(\sqrt[3]{x + 1}) \\[3ex] = (\sqrt[3]{x + 1})^3 + 1 \\[3ex] = x + 1 + 1 \\[3ex] = x + 2 \\[3ex] \ne x \\[3ex]$ Neither condition is satisfied
Both are not inverses of each other.

Who wants to do more work?
May we actually find the inverse of each function?
Let us find the inverse of each function.

$f(x) = \sqrt[3]{x + 1} \\[3ex] y = \sqrt[3]{x + 1} \\[3ex] Interchange\:\: x\:\: and\:\: y \\[3ex] x = \sqrt[3]{y + 1} \\[3ex] Solve\:\: for\:\: y \\[3ex] \sqrt[3]{y + 1} = x \\[3ex] Cube\;\;both\;\;sides \\[3ex] y + 1 = x^3 \\[3ex] y = x^3 - 1\\[3ex] \therefore f^{-1}(x) = x^3 - 1 \\[3ex] But: \\[3ex] g(x) = x^3 + 1 \\[3ex] \implies g(x) \ne f^{-1}(x) \\[3ex]$ $\therefore f(x)$ and $g(x)$ are not inverse functions

$g(x) = x^3 + 1 \\[3ex] y = x^3 + 1 \\[3ex] Interchange\:\: x\:\: and\:\: y \\[3ex] x = y^3 + 1 \\[3ex] Solve\:\: for\:\: y \\[3ex] y^3 + 1 = x \\[3ex] y^3 = x - 1 \\[3ex] Take\;\;the\;\;cube\;\;root\;\;of\;\;both\;\;sides \\[3ex] y = \sqrt[3]{x - 1} \\[3ex] \therefore g^{-1}(x) = \sqrt[3]{x - 1} \\[3ex] But: \\[3ex] f(x) = \sqrt[3]{x + 1} \\[3ex] \implies f(x) \ne g^{-1}(x) \\[3ex]$ $\therefore g(x)$ and $f(x)$ are not inverse functions
(3.)

(4.) Determine $f^{-1}(x)$ for $f(x) = 7 + 12x$

$f(x) = 7 + 12x \\[3ex] y = 7 + 12x \\[3ex] Interchange\:\: x\:\: and\:\: y \\[3ex] x = 7 + 12y \\[3ex] Solve\:\: for\:\: y \\[3ex] 7 + 12y = x \\[3ex] 12y = x - 7 \\[3ex] y = \dfrac{x - 7}{12} \\[5ex] \therefore f^{-1}(x) = \dfrac{x - 7}{12} \\[5ex]$ Test for Inverses

$(f \circ f^{-1})(x) \\[3ex] = f(f^{-1}(x)) \\[3ex] = f\left(\dfrac{x - 7}{12}\right) \\[5ex] But\:\: f(x) = 7 + 12x \\[5ex] So\:\: f\left(\dfrac{x - 7}{12}\right) = 7 + 12\left(\dfrac{x - 7}{12}\right) \\[5ex] = 7 + 1(x - 7) \\[3ex] = 7 + x - 7 \\[5ex] = x \\[7ex] (f^{-1} \circ f)(x) \\[3ex] = f^{-1}(f(x)) \\[3ex] = f^{-1}(7 + 12x) \\[3ex] But\:\: f^{-1}(x) = \dfrac{x - 7}{12} \\[5ex] So\:\: f^{-1}(7 + 12x) = \dfrac{(7 + 12x) - 7}{12} \\[5ex] = \dfrac{7 + 12x - 7}{12} \\[5ex] = \dfrac{12x}{12} \\[5ex] = x \\[3ex] (f \circ f^{-1})(x) = (f^{-1} \circ f)(x) = x$
(5.)

(6.)

(7.)

(8.)