Question

Match the one-to-one functions with their inverse functions.

Answer 1

I'll match them for you, but to find the inverse of an equation, all you must do is

1. Switch x and y
2. Solve for y again for the "inverse" !

$f(x)^{-1} = 5x$  →  $f(x) = \frac{x}{5}$

$f(x)^{-1} = \frac{x^{3}}{2}$  →  $f(x) = \sqrt[3]{2x}$

$f(x)^{-1} = x + 10$  →  $f(x) = x - 10$

$f(x)^{-1} = \frac{3(x+17)}{2}$  →  $f(x) = \frac{2x}{3} -17$

Hope I help ! :)

Answer 2

ANSWER

$\boxed {f(x)= \frac{2x}{3} - 17\to \: f ^{ - 1} (x)=\frac{3x + 51}{2}}$

$\boxed {f(x) = x - 10 \to {f}^{ - 1} (x) = x + 10 }$

$\boxed {f(x) = \sqrt[3]{2x} \to {f}^{ - 1} (x) = \frac{ {x}^{3} }{2} }$

$\boxed {f(x) = \frac{x}{5} \to{f}^{ - 1} (x) = 5x}$

EXPLANATION

1.

Given :

$f(x) = \frac{2x}{3} - 17$

Let

$y =\frac{2x}{3} - 17$

Interchange x and y.

$x=\frac{2y}{3} - 17$

Solve for y.

$x + 17=\frac{2y}{3}$

$3x + 51=2y$

$y=\frac{3x + 51}{2}$

$f ^{ - 1} (x)=\frac{3x + 51}{2}$

2.

Given: f(x)=x-10

Let y=x-10

Interchange x and y.

x=y-10

Solve for y.

y=x+10

This implies that,

${f}^{ - 1} (x) = x + 10$

3.

Given:

$f(x) = \sqrt[3]{2x}$

Let

$y=\sqrt[3]{2x}$

Interchange x and y.

$x=\sqrt[3]{2y}$

solve for y.

${x}^{3} = 2y$

$y = \frac{ {x}^{3} }{2}$

${f}^{ - 1} (x) = \frac{ {x}^{3} }{2}$

4.

Given:

$f(x) = \frac{x}{5}$

Let

$y = \frac{x}{5}$

Interchange x and y.

$x = \frac{y}{5}$

Solve for y.

$y = 5x$

${f}^{ - 1} (x) = 5x$