If $f(0)=f(2)=0$ and $f''>0$ on $(0,3)$ then $f(1) < 0$

Question

I'm trying to prove the question below and I think using the property/definition of convexity would be the easiest way. Can anyone tell of another way or confirm my way is adequate? I feel this could be so simple I am making an error.

Suppose $f$ is twice differentiable on $(-1, 3)$ and $f''(x) > 0$ on $(0, 3)$. Suppose $f(0) = 0, f(2) = 0.$ Prove $f(1) < 0$.

Since $f''(x) > 0$ then $f$ is strictly convex on $[0, 3]$. Thus, we know for $0 \leq x_1 < x_2 < x_3 \leq 3$ we know the following inequality to be true:

$(x_3 -x_1)f(x_2) < (x_3 - x_2)f(x_1) + (x_2 -x_1)f(x_3)$. Taking $x_1 = 0, x_2 = 1, x_3 = 2$ we get:

$2f(1) < f(0) + f(2) \rightarrow f(1) < 0.$

Answer 1

Note that $f''$ is positive on $(0, 3)$ so that $f'$ is strictly increasing on $(0, 3)$. And since $f(0) = f(2) = 0$ it follows that $f'(c) = 0$ for some $c \in (0, 2)$. Because of strict monotone nature we have $f'(x) < 0$ for $x \in (0, c)$ and $f'(x) > 0$ for all $x \in (c, 2)$. So $f$ is strictly decreasing in $[0, c]$ and strictly increasing in $[c, 2]$. It follows that $f(x) < 0$ for all $x \in (0, 2)$ and hence $f(1)$ in particular is negative.

Answer 2

I think it's a bit convoluted. Remember that strict convexity on $[a,b]$ means that for all $t\in(0,1)$ we have

$$f(t\cdot a+(1-t)\cdot b)<t\cdot f(a)+(1-t)\cdot f(b)$$

Using $a=0$, $b=2$ and $t=1/2$ we get $f(1)<0$ as desired.