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=(10-t)/(2) \\ H^(-1)=\sqrt[3]{(10-t)/(2)}^{} \end{gathered}`

The inverse function is:

`H^(-1)=\sqrt[3]{(10-t)/(2)}^{}`

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