Question

Match each function with its inverse function. use function composition to determine your answers.

Answer 1

QUESTION 1

`\boxed {f(x) = 2x + 6 \to \: g(x) = (1)/(2)x - 3 }`

The reason is that:

`g(f(x)) = 2( (1)/(2) x - 3) + 6`

Expand:

`g(f(x)) = x - 6+ 6`

`g(f(x)) = x`

QUESTION 2

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

The reason that

`g(f(x)) = - (1)/(2) (3 - 2x - 3)`

`g(f(x)) = - (1)/(2) (- 2x )`

`g(f(x)) =x`

QUESTION 3

`\boxed {f(x) = \sqrt[3]{3x}+ 2 \to \: g(x) = \frac{ {(x - 3)}^(3) }{3} }`

The reason is that:

`f(g(x)) = \sqrt[3]{ \frac{3 {(x - 2)}^(3) }{3} } + 2`

`f(g(x)) = x-2 + 2`

`f(g(x))=x`

QUESTION 4

`\boxed {f(x)=3\sqrt[3]{x + 2} \to \: g(x) = (1)/(27) {x}^(3) - 2}`

The reason is that

`f(g(x)) = 3 \sqrt[3]{ (1)/(27) {x}^(3) - 2 + 2}`

`f(g(x)) = 3 \sqrt[3]{ (1)/(27) {x}^(3) }`

`f(g(x)) = 3 * (1)/(3) x`

`f(g(x)) =x`