In part a we must find the inverse of:
![f(x)=x^2-10,x\ge0](https://img.qammunity.org/2023/formulas/mathematics/college/mfmw4tqc06k54ikzn9xba5hewl5jt8qkqq.png)
For this purpose we'll need to make the following replacements:
![\text{We replace }f(x)\text{ with }x\text{ and }x\text{ with }f^(-1)(x)](https://img.qammunity.org/2023/formulas/mathematics/college/uek5pphv1cajzuwlbnrk2w7mla4qgqwydk.png)
With these replacements we get:
![f(x)=x^2-10\rightarrow x=(f^(-1)(x))^2-10](https://img.qammunity.org/2023/formulas/mathematics/college/yrmx0c0o5b3g44rxk7f461pseuspd3aw26.png)
So we need to solve this equation for f^(-1):
![x=(f^(-1)(x))^2-10](https://img.qammunity.org/2023/formulas/mathematics/college/sa0rhr6l5w7gab2jr0zic5n8kdd24s9hmq.png)
We add 10 at both sides of the equation:
![\begin{gathered} x=(f^(-1)(x))^2-10 \\ x+10=(f^(-1)(x))^2-10+10 \\ x+10=(f^(-1)(x))^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/mtpo9g1i7idy010jcd8w4u9z62plqsox3u.png)
Now we apply the square root at both sides:
![\begin{gathered} x+10=(f^(-1)(x))^2 \\ √(x+10)=\sqrt{(f^(-1)(x))^2} \\ f^(-1)(x)=\sqrt[]{x+10} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vzgd0u36vx4sbngfx505wux9eplxm9spuo.png)
So the answer to part a is:
![f^(-1)(x)=\sqrt[]{x+10}](https://img.qammunity.org/2023/formulas/mathematics/college/832kvczc7vf3z63auoal2vvbjmcphwa87n.png)
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