In part a we must find the inverse of:
[tex]f(x)=x^2-10,x\ge0[/tex]
For this purpose we'll need to make the following replacements:
[tex]\text{We replace }f(x)\text{ with }x\text{ and }x\text{ with }f^{-1}(x)[/tex]
With these replacements we get:
[tex]f(x)=x^2-10\rightarrow x=(f^{-1}(x))^2-10[/tex]
So we need to solve this equation for f^(-1):
[tex]x=(f^{-1}(x))^2-10[/tex]
We add 10 at both sides of the equation:
[tex]\begin{gathered} x=(f^{-1}(x))^2-10 \\ x+10=(f^{-1}(x))^2-10+10 \\ x+10=(f^{-1}(x))^2 \end{gathered}[/tex]
Now we apply the square root at both sides:
[tex]\begin{gathered} x+10=(f^{-1}(x))^2 \\ \sqrt{x+10}=\sqrt{(f^{-1}(x))^2} \\ f^{-1}(x)=\sqrt[]{x+10} \end{gathered}[/tex]
So the answer to part a is:
[tex]f^{-1}(x)=\sqrt[]{x+10}[/tex]
In part b we must find the domain and range of both f and f^(-1). The domain is composed of all the possible x values for which the function ha
