I want some verification to see if I am doing this question right. Here is the question, and my attempt:
Suppose $g(u) = \int_{2}^{u} \sqrt{1 + t^3} \ dt$. Find the value of $(g^{-1})'(0)$ if it exists.
So first I found the derivative of $g(u)$, which was $$g'(u) = \sqrt{1 + u^3}$$
Then I let $v = g'(u)$. The domain of $g'$ is $u\in[-1, \infty)$ and the range of $g'$ is $v\in[0,\infty)$. Now solve for $u$, so \begin{align*} v &= \sqrt{1 + u^3} \\ v^2 &= 1 + u^3 \\ u^3 &= v^2 - 1 \\ u &= \sqrt[3]{v^2 - 1} \end{align*}
So the inverse is $u = \sqrt[3]{v^2 - 1}$, or $(g^{-1})'(v) = \sqrt[3]{v^2 - 1}$. The domain of $(g^{-1})'$ is $v\in[0,\infty)$, and the range for $(g^{-1})'$ is $u\in[-1,\infty)$.
Now find $(g^{-1})'(0)$, so sub $v = 0$. \begin{align*} (g^{-1})'(v) &= \sqrt[3]{v^2 - 1} \\ (g^{-1})'(0) &= \sqrt[3]{0^2 - 1} \\ (g^{-1})'(0) &= \sqrt[3]{-1} \\ (g^{-1})'(0) &= -1 \\ \end{align*}
Since the value $-1$ is within the range, this value exists. Therefore, the value of $(g^-1)'(0)$ is $-1$.
This is my attempt, not sure if it is correct...Would like some suggestions or advice, if it is off somewhere.
You're not doing it correctly, you're calculating $(g')^{-1}$ instead of $(g^{-1})'$. To do it correctly, you need the formula $$ (g^{-1})'(x) = \frac{1}{g'(g^{-1}(x))}$$ Specifically $$ (g^{-1})'(0) = \frac{1}{g'(g^{-1}(0))}$$ You properly calculated $$g'(u) = \sqrt{1+u^3} $$ so you just need to find $g^{-1}(0)$ which is such a value $u$ for which $g(u) = 0$, that is $$ 0 = \int_{2}^u\sqrt{1+t^3}dt$$ even without solving the integral it can be noticed that this equation is satisfied for $u=2$. We have then
$$ (g^{-1})'(0) = \frac{1}{g'(g^{-1}(0))} = \frac{1}{g'(2)} = \frac{1}{\sqrt{1+2^3}} = \frac13$$