Question

Given H(t)=10-2t^2 , find h^-1 (t), is h (t) one to one ? What does this imply about h^-1(t)

Answer 1

We have the function:

$H(t)=10-5t^2$

And we have to find the inverse function.

We can write this as:

$H(H^{-1}(t))=t$

We can solve this as:

$\begin{gathered} H(H^{-1}(t))=10-2(H^{-1})^3=t \\ 10-2(H^{-1})^3=t \\ 2(H^{-1})^3=10-t \\ (H^{-1})^3=\frac{10-t}{2} \\ H^{-1}=\sqrt[3]{\frac{10-t}{2}}^{} \end{gathered}$

The inverse function is:

$H^{-1}=\sqrt[3]{\frac{10-t}{2}}^{}$

The domain of H is all the real numbers, as the domain of H^-1.