Tangent Lines- h(x)=f(g(x)) Assume $f$ and $g$ are differentiable on their domains with
$h(x)=f(g(x))$.
Suppose the equation of the line tangent to the graph
of $g$ at the point $(4,70)$ is $y=3x-5$ and the equation of the line
tangent to the graph of $f$ at $(7,9)$ is $y= -2x+23$
a. calculate $h(4)$ and $h'(4)$.
b. Determine an equation of the line tangent to the graph of $h$ at
the point of the graph where $x=4$.

I'm lost on a, can anyone give me a hint on how to approach it?
We have insufficient information about $f(x)$ to solve part a. All we know about $f(x)$ is that it passes through $(7,9)$ and that it has derivative $f'(7)=-2$. This could be satisfied by $f(x)=-2x+23$, in which case $h(4)$ would be $-117$ but it could also be satisfied by $f(x)=x^2-16x+72$, in which case, $h(4)$ would be $3852$. Presumably the value of $70$ was a typo and should have been $7$, in which case $h(4)$ would have been: $4\underbrace{\rightarrow}_{g}7\underbrace{\rightarrow}_{f}\underline{9}$.

For part b, it suffices to use the chain rule.
Hint : $h(x)=f(3x-5)$

Also $dh/dx=f'(g(x)) g'(x)$