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) = (x)/(5)`

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

`f(x)^(-1) = x + 10` →
`f(x) = x - 10`

`f(x)^(-1) = (3(x+17))/(2)` →
`f(x) = (2x)/(3) -17`

Hope I help ! :)

Answer 2

`\boxed {f(x)= (2x)/(3) - 17\to \: f ^( - 1) (x)=(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) = (x)/(5) \to{f}^( - 1) (x) = 5x}`

Step-by-step explanation

1.

Given :

`f(x) = (2x)/(3) - 17`

Let

`y =(2x)/(3) - 17`

Interchange x and y.

`x=(2y)/(3) - 17`

Solve for y.

`x + 17=(2y)/(3)`

`3x + 51=2y`

`y=(3x + 51)/(2)`

`f ^( - 1) (x)=(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) = (x)/(5)`

Let

`y = (x)/(5)`

Interchange x and y.

`x = (y)/(5)`

Solve for y.

`y = 5x`

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