Question

If _______ , what is f^-1(x)? f–1(x) = 9x + 18 ________ f–1(x) = 9x + 2 ________

Answer 1

we have

$f(x)=\frac{1}{9}x-2$

Let

$y=f(x)$

$y=\frac{1}{9}x-2$

To find the inverse

Exchanges the variables x for y and y for x

$x=\frac{1}{9}y-2$

Isolate the variable y

Multiply by $9$ both sides

$9x=y-18$

Adds $18$ both sides

$9x+18=y-18+18$

$y=9x+18$

Let

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

$f^{-1}(x)=9x+18$ -------> the inverse function

therefore

the answer is

$f^{-1}(x)=9x+18$

Answer 2

If f(x)=1/9x-2
then f^-1(x) will be
lets assume f(x)=y
y=1/9x-2
y+2=1/9x
9(y+2)=x
f^-1(x)=9(y+2)

rest u can do with the same method