Question

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = ln(x2 + 7x + 14), [−4, 1]

Answer 1

$f(x)=\ln(x^2+7x+14)\implies f'(x)=\dfrac{2x+7}{x^2+7x+14}$

Note that $x^2+7x+14>0$ for all $x\in\mathbb R$, so that $f(x)$ is defined everywhere and $f'(x)$ is continuous. This means the only critical points, if any, that occur for $f(x)$ occur when $f'(x)=0$:

$\dfrac{2x+7}{x^2+7x+14}=0\implies2x+7=0\implies x=-\dfrac72$

Now, the second derivative is

$f''(x)=\dfrac{2(x^2+7x+14)-(2x+7)^2}{(x^2+7x+14)^2}=-\dfrac{2x^2+14x+21}{(x^2+7x+14)^2}$

and at the critical point, we have

$f''\left(-\dfrac72\right)=\dfrac87>0$

which means that a minimum occurs at the point $\left(-\dfrac72,\ln\dfrac74\right)\approx(-3.5,0.5596)$.

At the endpoints of the given interval, we have

$f(-4)=\ln2\approx0.6931$
$f(1)=\ln22\approx3.091$

So, on the given interval, there is an absolute maximum at $x=1$ and an absolute minimum at $x=-\dfrac72$.