what does $f''$ suggests about the graph of $f$?

Question

As I read the question, I only understand $g'(0)=0$ means $g$ has a local minimum or maximum there, so the solution must be A,B or E.

That's all. Can you help me figure out what $g''$ suggests?

Thank you

Answer 1

• $g'(0) = 0$ eliminates (C), since it's slope of tangent at $x=0$ is positive.
• $g''(-1) > 0$ indicates the local convexity at $x=-1$, so (E) is eliminated.
• $g''(x) < 0$ on $x \in (0,2)$ indicates the concavity there, so (B) and (D) are eliminated.

Observe that (A) verifies the above three criteria, so (A) is the answer.

Answer 2

$g''(x)>0$ tells you that it is strictly convex at x. Here is why:

Let $x<y$

$0<\int\limits_x^yf(t)dt=f'(y)-f'(x)$

so $(f'(y)-f'(x))(y-x)>0$ which is an alternative definition of convex