Question

Let f(x) = x^2 – 16. Find f^–1(x).

Answer 1

anything to a negative power can be written with a positive power if it is moved to the denominator [x^(-2)]=1/(x^2)
what do you think f^(-1)(x) is?

Answer 2

is it 1/x^2-16?

Answer 3

yes, 1/(x^2-16)

Answer 4

haha no that is not what they are looking for.
\[f^{-1}(x) \] refers to the inverse function
i hate the notation because it does look like an exponent, however f is not a variable its a function

Answer 5

basically you have to solve for x here.
let f(x) be y

\[y = x^{2} -16\]
\[x = \pm \sqrt{y+16}\]
\[\rightarrow f^{-1}(x) = \sqrt{x+16}\]
important note, range of f(x) is y>-16
so domain of inverse must be x>-16