Showing that there exists $c \in (a, b)$ with $f''(c) > 0$. Suppose we are given that a function, say $f$ is twice differentiable on $(a, b)$. Now $x_1, x_2, x_3 \in (a,b)$ and $f(x_2) &lt; f(x_1)$ and $f(x_2) &lt; f(x_3)$ (also $x_1 &lt; x_2 &lt; x_3$). I want to show that there is a $c \in (a, b)$ with $f''(c) &gt; 0$. I've heard that we can apply the MVT three times here, but I do not see where that would help.

Graphically, this is intuitive and my belief is that there are many such $c$'s in $(a, b)$ such that $f''(c) &gt; 0$. In fact, every $x  \in (a,b)$ has $f''(x) &gt; 0.$

Anyways the mean value theorem states that: Suppose that $a , b \in \mathbb{R}$ with $a &lt; b$. If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ then there exists a $c \in (a, b)$ such that $f(b) - f(a) = f'(c)(b-a)$. 

Can you give me some hint?
First we apply the MVT twice

$$f(x_1) - f(x_2) = f'(c_1)(x_1-x_2),$$ so $f'(c_1) &lt; 0$ for $c_1 \in (x_1,x_2)$.

$$f(x_3)-f(x_2) = f'(c_2)(x_3-x_2),$$ so $f'(c_2) &gt; 0$ for $c_2 \in (x_2,x_3)$.

Now, $c_2 &gt; c_1$, so $$f'(c_2)-f'(c_1) = f''(c)(c_2-c_1)$$ since the lhs is positive, and since $c_2 &gt;c_1$ this $c$ gives $f''(c)&gt;0$.

Notice that in all of this, we only look at the signs of things, and this is enough without worrying about any particular values.
First we apply the MVT twice

$$f(x_1) - f(x_2) = f'(c_1)(x_1-x_2),$$ so $f'(c_1) &lt; 0$ for $c_1 \in (x_1,x_2)$.

$$f(x_3)-f(x_2) = f'(c_2)(x_3-x_2),$$ so $f'(c_2) &gt; 0$ for $c_2 \in (x_2,x_3)$.

Now, $c_2 &gt; c_1$, so $$f'(c_2)-f'(c_1) = f''(c)(c_2-c_1)$$ since the lhs is positive, and since $c_2 &gt;c_1$ this $c$ gives $f''(c)&gt;0$.

Notice that in all of this, we only look at the signs of things, and this is enough without worrying about any particular values.