Question

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

Answer 1

QUESTION 1

The given function is

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

To find the inverse function, we let

$y = f(x)$

This implies that,

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

This will give us,

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

We multiply through by 3 to obtain;

$3(y + 17) = 2x$

We now interchange x and y to obtain,

$3x = 2y - 51$

We make y the subject to obtain,

$3(x + 17) = 2y$

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

This implies that,

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

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

QUESTION 2

The given function is

$f(x) = x - 10$

To find the inverse function we let
$y = x - 10$

We then interchange x and y to obtain,

$x = y - 10$

We solve for y to obtain,

$y = x + 10$

Therefore the inverse function is

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

Hence,

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

QUESTION 3.

The given function is

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

We want to find the inverse so we let

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

We now interchange x and y to obtain,

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

We now make y the subject, by first taking the cube of both sides of the equation.

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

Divide through by 2 to get,

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

Or

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

This implies that,

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

Therefore

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

QUESTION 4

The given function is

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

We let

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

Interchange x and y to get,

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

Make y the subject to get,

$y = 5x$

This implies that,

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

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

Answer 2

F(x) = x/5
F(x)= 3^SQR 2x
F(x) =x - 10
F(x) = 2x/3 - 17

These would be your answers in order :)