Proofs of Derivative Applications Facts — Question 7

Question 7 - Solution

To show that $f$ is concave down on $(a,b)$, we must show that the derivative $f'(x)$ is decreasing on $(a,b)$.

Let $x_1,x_2\in(a,b)$ with $x_1<x_2$. Since $f$ is twice differentiable on $(a,b)$, the function $f'$ is differentiable on $(a,b)$ and therefore continuous on $[x_1,x_2]$.

Apply the Mean Value Theorem to $f'$ on the interval $[x_1,x_2]$. There exists a number $c\in(x_1,x_2)$ such that $$f''(c)=\frac{f'(x_2)-f'(x_1)}{x_2-x_1}.$$

By assumption, $f''(c)<0$, and since $x_2-x_1>0$, it follows that $$\frac{f'(x_2)-f'(x_1)}{x_2-x_1}<0.$$

Multiply both sides by $x_2-x_1$: $$f'(x_2)-f'(x_1)<0.$$

Thus, $$f'(x_2)<f'(x_1).$$

Since this holds for all $x_1<x_2$ in $(a,b)$, the derivative $f'(x)$ is decreasing on $(a,b)$.

Therefore, the function $f$ is concave down on $(a,b)$.

$$\boxed{f\text{ is concave down on }(a,b)}$$