Verify algebraically that $f(f^{-1}(x)) = x$ for $f(x) = \sqrt{x}-5$
So i started going through this question, and I found the inverse as $(x+5)^2$, but the part I'm having trouble with is verifying that the inverse function that I found is equal to "x" like the question asked. There wasn't much in the curriculum or in my textbook on explaining what that means or how to do it. Wondering if someone out there can help explain how to verify that the inverse function equals "x" please and thanks.
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$\begingroup$Given $f(\color{green}{x})=\sqrt {\color{green}{x}}-5, \color {blue}{f^{-1}(x)=(x+5)^2}$, it follows that
\begin{align}f(\color{blue}{f^{-1}(x)})=f(\color{blue}{(x+5)^2})\end{align}\begin{align}f(\color{green}{(x+5)^2})=\sqrt{\color{green}{(x+5)^2}}-5=|x+5|-5=x (x\geq 0)\end{align}
\begin{align}f(\color{blue}{f^{-1}(x)})=x\end{align}
$\endgroup$ 3 $\begingroup$For simplicity, consider $t=f^{-1}(x)$ i.e. $t=(x+5)^2$. Now, you have to prove that $f\big(f^{-1}(x)\big)=x$ i.e. $f(t)=x$.
Now, just put the value of $t$. If the result on simplification comes out to be $x$, you are done!
Hope this helps. Ask anything if not clear.
$\endgroup$ $\begingroup$So you have $f(x)=\sqrt{x}-5,f^{-1}(x)=(x+5)^2$, and you want to find out what $f(f^{-1}(x))$ is. To get you comfortable with the notation:
$$\begin{align}f(\color{green}{x})&=\sqrt{\color{green}{x}}-5\\f(1+1)&=\sqrt{1+1}-5\\f(x^2+7)&=\sqrt{x^2+7}-5\\f(\blacksquare)&=\sqrt{\blacksquare}-5\\f(f(x))&=\sqrt{f(x)}-5\end{align}$$ Can you see how to find $f(f^{-1}(x))$?
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