Derivative sketching

Question

I have been able to do a few of these however this one has me stumped. I can't think of any function to model that I can use as reference and I have no idea what is supposed to happen at x=-0.4. I assume a saddle point/inflexion point?

Also, I'd rather not have just an answer. If someone could walk through their process in an informative manner I would appreciate it.

Answer 1

First you know at all blue points $f'=0$ and because $f'$ intersects your $y$-axis at all blue points, monotony changes look here f.e. http://sites.csn.edu/ehutchinson/notes/4_3notes181.pdf

and therefore you know these are lokal extrema. Ok so far?

Answer 2

It's very simple if you recall a few facts about the derivative of a function:

1. a function has a stationary point where its derivative is zero
2. a function is increasing where its derivative is positive
3. a function is decreasing where its derivative is negative

So, start from the stationary points. We see that they are at $x = -1.1, 0.3$ and $6$. Draw a dash in a blank cartesian plane for each of these. It doesn't really matter if you draw them in the positive $y$ plane or not, we can change that later.

Now, starting from $-\infty$ look at the sign of $f'(x)$. Initially it's negative, so for $x < -1.1$, $f(x)$ will be decreasing. You can draw a line that starts high and descends to the dash that you made in correspondence of $x = -1.1$. Then $f'(x)$ is positive in $(-1.1, 0.3)$, so in this interval $f(x)$ will be always increasing. From this we deduce that actually the second dash has to be higher than the first.

Continue in this way until you have exhausted the $x$ axis.

Note: at $x = -0.4$, as you correctly said, there is an inflection point, since the function keeps increasing both at its right and its left.