Finding the marked values of x on a graph 
I would assume that since $x_3$ is the local maximum(or absolute maximum) on the graph of $f$ prime, that it would be the greatest on the graph of $f.$ 
However, this problem is online, and in order for me to get it right, I have to get all of the problems shown right. Would it be right to see the graph at local and absolute extrema? If not, how would I go about solving a problem like this?
 A: This is a bit of a tricky problem because the graph of $f'$ naturally misleads our intuitions about $f$ -- we're used to looking at graphs of functions, not of their derivatives!
A. It's $x6$. Note that the $f'(x) > 0$ for all $x$ on that graph. You know that a function $f$ is increasing if $f'(x) > 0$. So $f(x)$ was increasing for all $x$. Since $f(x)$ was increasing for all $x$, we have that $f(x)$ must be largest for the largest $x$, which is $x6$.
This is really important. The fluctuations in $f'(x)$ meant that the function $f$ was just `growing' at different rates, but that rate was always greater than $0$, so $f$ was always increasing. Even though there was some fluctuation in the derivative, the derivative never hit $f'(x) = 0$, which would've meant a maximum or minimum.
B. By reasoning similar to the above, it's $x1$. The function $f(x)$ is increasing on all $x$, so the smallest value of $f(x)$ must be attained at the smallest $x$, which is $x1$.
C. Right. You're looking at a graph of $f'(x)$, so just take the greatest marked point.
D. You're looking at a graph of $f'(x)$. At which marked point does $f'(x)$ take the smallest value? It's $x5$.
E. At $x3$, $f''(x) = 0$. You know this because you have a graph of $f'(x)$, and that's where $f'(x)$ has a local maximum. The derivative of a function $f$ at a local maximum or minimum $x$ is $0$. By similar reasoning, it can't be $x5$ either. If you compare the slopes at all the points of $f'(x)$, you can see that $x6$ has the steepest upward-slope by far. So that's where the derivative of $f'(x)$  -- $f''(x)$ must be the greatest.
F. Wrong. Again, look at the slopes at all the points of $f'(x)$. You can see that $x4$ is the only point at which the slope/derivative of $f'(x)$ is negative, so that's where the derivative of $f''(x)$ is smallest.
