Below is the graph of the derivative of a function $f$ which is continuous on all real numbers. Which one of the followings are always correct?

Question

Below is the graph of the derivative of a function $f$ which is continuous on all real numbers

Which one of the followings are always correct?

I. $f(2)\lt f(4).$

II. The function $f$ has a local minimum at $x=3$.

III. The function $f$ is increasing on the interval $(-\infty,-1)$

I can figure out the last two options:

Firstly, third option is wrong because we see that the derivative of this function changes from negative to positive at a point smaller than $-1$, so the function is not always increasing on $(-\infty,-1)$.

Second option is correct because the derivative changes from negative to positive at the point $x=3$, meaning $f$ has a minimum point at 3.

Now for the first option, I see that the function is decreasing at $2$ and increasing on $4$, but I believe this does not necessarily mean that $f(2)\lt f(4)$.

How do we explain that $f(2)\lt f(4)$ will always be true?

Answer 1

The question sounds like it's trying to indicate that only one of the options is correct, and II is easily seen to be correct, by the first derivative test.

That said, using integration, we can see that I is correct as well. Recall that, by the fundamental theorem of calculus, $$f(4) - f(2) = f(3) - f(2) + f(4) - f(3) = \int_2^3 f'(x) \, \mathrm{d}x + \int_3^4 f'(x) \, \mathrm{d}x.$$

Visually, it seems that, for $2 \le x < 3$, we have that $f'(x) \ge -2$. For $x > 3$, we seem to have $f'(x) \ge 2$. Thus, $$f(4) - f(2) = \int_2^3 f'(x) \, \mathrm{d}x + \int_3^4 f'(x) \, \mathrm{d}x \ge \int_2^3 -2 \, \mathrm{d}x + \int_3^4 2 \, \mathrm{d}x = 0.$$ Moreover, the non-zero slope on the line forming $f'(x)$ past $x = 3$ tells us that the inequality is strict. Thus, $f(4) > f(2)$.