Question

PLEASE HELP IM SUPER BEHIND GEOMETRY 25 PTS

Answer 1

1)
`\therefore f^(-1)(x)=(1)/(2)x^2-3,\:x\ge0`.

2)
`f^(-1)(x)=\sqrt[4]{x-16},\:x\ge16`

3)
`f^(-1)(x)=(8)/(3)x-(8)/(3)`

Explanation:

1) The given function is
`f(x)=√(2x=6),\:\:x\ge-3`.

Let
`y=√(2x-6)`.

Interchange x and y to get:

`x=√(2y+6)`.

Square both sides;

`x^2=2y+6`.

Solve for y.

`x^2-6=2y`.

`(1)/(2)x^2-3=y`.

`\therefore f^(-1)(x)=(1)/(2)x^2-3,\:\:x\ge0`.

2) The given function is
`f(x)=x^4+16,\:\:x\ge0`

Let
`y=x^4+16`

Interchange x and y;

`x=y^4+16`

Solve for y

`x-16=y^4`

`\sqrt[4]{x-16}=y`,

Therefore the inverse is

`f^(-1)(x)=\sqrt[4]{x-16},\:x\ge16`

3) The given function is
`f(x)=(3x)/(8)+1`

Let
`y=(3x)/(8)+1`

Interchange x and y

`x=(3y)/(8)+1`

Solve for y:

`x-1=(3y)/(8)`

`8x-8=3y`

`(8)/(3)x-(8)/(3)=y`

Therefore the inverse is

`f^(-1)(x)=(8)/(3)x-(8)/(3)`