$f'(5) = \lim_{h\to 0} \frac{2(6+h)^2-72}{ h} $ What is $f(x)$? I'm not sure how to solve this. 
$$
f'(5) = \lim_{h\to 0} \frac{2(6+h)^2 - 72}{h}
$$
It is $f$ prime of $5$ above. Solve for f(x). If someone could please explain the process of solving and not just give an answer, that would be most helpful. 
Thank you!
 A: It is the derivative of $f(x) = 2(x+1)^2$ at $x = 5$. Thus:
$f(5+h) = 2(6+h)^2$, and $f(5) = 2\cdot 6^2 = 72$. Thus:
$f'(5) = \displaystyle \lim_{h \to 0} \dfrac{f(5+h) - f(5)}{h} = \displaystyle \lim_{h \to 0} \dfrac{2(6+h)^2 - 72}{h} = .... = 24$.
My original answer was that $f(x) = 2x^2$ at $x = 6$. Both gives the same answer $24$, but of course "the right function" is $f(x) = 2(x+1)^2$.
A: We obviously want to morph that limit into something of the form
\begin{equation}
\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}
\end{equation}
Luckily, in the question you provided, everything is already in the correct form - that is, you can easily see that $f(x)$ is something like $2(x)^2$.
If we were to use that back-of-the-envelope guess and put it in the limit-equation that I first wrote, we would have
$$\lim_{h\rightarrow0} \frac{2(x+h)^2-2x^2}{h}$$
And if you were to plug in $x=6$, it would essentially be what you wrote in your original question. However, that would correspond to $f'(6)$, and we want it to be $f'(5)$. To fix that, simply change $f(x)$ to $2(x+1)^2$. That would then change the limit-equation to
$$\lim_{h\rightarrow0} \frac{2(x+1+h)^2-2(x+1)^2}{h}$$
Setting $x=5$, we arrive at
$$\lim_{h\rightarrow0} \frac{2(6+h)^2-2(6)^2}{h} = \lim_{h\rightarrow0} \frac{2(6+h)^2-72}{h} = f'(5)$$
So to answer you primary question,
$$\boxed{f(x) = 2(x+1)^2}$$
