# 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?

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?

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, $$f(4) > f(2)$$.
• Thanks for the answer. Can you please elaborate on how could you plug in $-2$ into $f'(x)$ in $\int_2^3 f'(x) \, \mathrm{d}x$? Shouldn't $f'(x)$ be a function? Mar 24, 2019 at 8:15
• Recall that, if $f(x) \ge g(x)$ for all $x \in (a, b)$, then $\int_a^b f(x) \, \mathrm{d}x \ge \int_a^b g(x) \, \mathrm{d}x$. Here, I'm replacing $f(x)$ with $f'(x)$ and $g(x)$ with the constant function $-2$. Mar 24, 2019 at 8:39