How to determine the derivative of $ f $ at $ x=2$ by looking at the graph only? How to determine the derivative of $ f $ at $ x=2$ (i.e., $ f^\prime(2) $) by looking at the graph only ?
I am well aware of the theory of the derivative and how to compute it. But how can I determine the exact derivative at $ x=2$ here only by looking at the graph ?
 A: This function seems like a cubic.
The roots are $-4$, $0$ and $2$, so
$$f(x)=kx(x+4)(x-2)=k(x^3+2x^2-8x)$$
Since $f(1)=-5k\approx -1.2$, then $k\approx 0.25=\frac14$.
A: It is not possible to get an exact answer without knowing the function, and determining the function is not the point of this problem.
The point of this problem is to get a reasonable estimate of $f'(2)$, based on the shape of the graph.  As a first approximation, the function is increasing, so $f'(2)$ should be positive.  Next, consider the points on the graph that are approximately $(1.5,-1)$ and $(2.25,1)$.  The line between these two points has slope $\frac{1-(-1)}{2.25-1.5}=\frac{2}{0.75}=\frac{8}{3}\approx 2.7$.  This line is approximately parallel to the function at $2$, so a fairly good approximation for $f'(2)$ is $2.7$.  If you answered $3$ or $2.5$ those are also reasonable answers.  If you answered $10$ that would be too high.
A: Unless you can infer an exact formula from that diagram, I don't think you can state an exact answer with certainty. But you can get a reasonable approximation by placing a straightedge tangent to the point where the graph passes through $x=2$ (or at least as close to "tangent" as you can make it by the imprecise method of placing an actual straightedge on an actual drawing and trying to estimate tangency by eye). You can then use the grid lines to estimate the slope of the straightedge.
On the other hand, given the context in which you found the problem, you might try assuming that the function has a simple formula that would be easy to differentiate, and then try to guess the formula. You have gotten at least two answers that make this assumption, from which they derive exact cubic functions that you can easily differentiate at $2$.
A: As you know, the exact result is
$$f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}.$$
So, by taking small enough values of $h$, you should be able to give a pretty good guess at what the answer is. Just looking at the graph, the author probably wants you to take $h = 1$ which gives a result of $5 - 0 = 5$ for the derivative. You could have also taken $h = -1$ and the result would be something like $1,25$, which is of course very different from $5$ but just as good an answer. Taking values of $h$ that are closer to $0$ will give you results that are closer to the actual value of $f'(2)$. Besides forcing you to go through that bit of logic, the exercise seems pretty bad.
A: If you draw a line tangent to it at the point $f(2)$ you can then get the equation of this line and since it's a straight line you can directly see the slope. In your cas if you draw a line approximately from $(0, -3)$ and  $(4, 5)$ you have line which seems to be pretty tangent. The accompanying equation of this line is $g(x) = 2x - 3$ Where you can directly get the slope which is 2. 
