Find the function $h(x) = g(2g^{-1}(x))$

Question

Show that the function $g(x) = x^4 + x^3 + 1$ is one-to-one on [0, 2]. In addition, for the function $h(x) = g(2g^{-1}(x))$, find h′(3).

For the first part, I manage to prove that g(x) is increasing using mean value theorem where $f'(c)=\frac{f(2)-f(1)}{2-1}>0$

However I'm stuck at the second part. I know that $g(g^{-1}(x))=x$ but I have no idea what to do with the 2 in $g(2g^{-1}(x))$.

Answer 1

Use the chain rule and also the inverse function rule. (You need to extend the proof of part 1 to [0,3] to show that the chain rule is valid).

Answer 2

I don't think the function $ h $ is well-defined in $ x = 3 $, since $ g $ takes the value $ g(x) = 3 $ for two distinct values of $ x $; if we restrict our attention to $ x \ge 0 $, it's easily seen that $ g^{-1}(3) = 1 $.

So, apply the chain rule to get: $$ \frac{\mathrm{d}h}{\mathrm{d}y}(3) = \frac{\mathrm{d}g}{\mathrm{d}x}(2g^{-1}(3)) \frac{\mathrm{d}(2g^{-1})}{\mathrm{d}y}(3) = \frac{\mathrm{d}g}{\mathrm{d}x}(2g^{-1}(3)) \frac{2}{\frac{\mathrm{d}g}{\mathrm{d}x}(g^{-1}(3))} = \frac{\mathrm{d}g}{\mathrm{d}x}(2) \frac{2}{\frac{\mathrm{d}g}{\mathrm{d}x}(1)} = \frac{88}{7}. $$