Find the graph of the derivative

Question

The graph of a function $f$ defined on $(0,2)$ with the property $\lim_{x\to 0^+}f(x)=\lim_{x\to 2^-}f(x)=\infty$ is as follows:

Answer 1

First, as I said in the comments you can see how this isn't a parabola since all parabolas don't have vertical asymptotes, whereas this function does.

Now for a general problem like this I will outline what I do. I would begin looking at the increasingness and decreasingness (I establish these non-words as words!) of the function. By this, I mean imagine tangent lines along the function. Look at the slope of these tangent lines, what do you notice about them? Now I'm sure you know that the derivative evaluated at a point is the slope of the tangent line at that point. So now if you notice that a tangent line's slope is negative the value of the derivative must be negative, likewise positive. A quick way to see if a slope is going to be negative or not is to look at whether or not it's a / (solidus) or \ (reverse solidus). If the slope looks like / then the slope is positive and \ it's negative (a proof of this is fairly straight forward and I can list it if you want). From this visual method one can also notice how negative the slope is. That is if the slope is more vertical at a given point and looks like \ as compared to another point then it must be more negative. Likewise if a slope is more vertical at one point and looks like / as compared to another point then the derivative must be more positive at that point.

Hopefully that helped for a general problem like this (personally I'm a fairly visual person so doing this process has become second nature to me)

Now let's look at this specific problem:

Near $0$ we see that the slope is very vertical and looks like \ so the derivative must be very negative here. Continuing towards $1$ the slope is still \ but continues to get less and less vertical so the derivative must be getting less and less negative* until it hits the point where the slope is actually flat which means that the slope is $0$ so the derivative is $0$. Continuing the slope becomes more and more positive until it's /very/ positive around $2$. In the end patching this information together the graph must go from very negative around $x=0$ to zero at $x=1$ to very positive at $x=2$. The only graph with these properties is D).

*Note that less negative means it is increasing, but still negative

Answer 2

The answer is $D)$, since the slope asymptotes to $\infty$ as $x\to0$ and $x\to2$, and there is a minimum at $x=1$. Also, the graph of $f$ is concave up, and hence, $f^\prime$ needs to have positive slope.