Question

All I need is someone to work this question out with me.. FAN AND MEDAL WILL BE AWARDED <3

Answer 1

\[f(x)=x^2-81\]

\[f(x)^{-1}=\frac{ 1 }{ x^2-81 }\]

Answer 2

okay

Answer 3

Thank you all but I'm going to have to go with @Nurali's answer. It seems way more legitimate. Thank you @Nurali for helping me through the problem :)

Answer 4

as already suggested, to get the inverse function or \(f^{-1}(x)\)
you simply do a quick switch of the variables like \(\bf f(x) = x^2 - 81 \implies \color{red}{y}= \color{blue}{x}^2 - 81\\
f^{-1}(x) \implies \color{blue}{x}= \color{red}{y}^2 - 81\)

Answer 5

keep in mind that Nurali is correct \(\bf f(x)^{-1} = \cfrac{1}{f(x)}\\
\textit{however } f(x)^{-1} \ne f^{-1}(x)\)

Answer 6

f(x)=x^2-81
y=x^2-81
x=y^2-81
y^2=x+81
y=sqrt(x+81)
f^-1(x) =sqrt(x+81)

Answer 7

if what it's asked is the RECIPROCAL, then yes, it's \(\cfrac{1}{f(x)}\)
if what's asked is the INVERSE, then it's as shown above, by swithching the variables

Answer 8

Sorry i forgot @LaynaMae @jdoe0001

Answer 9

well, tis ok, the notation is kinda ambiguous