Science:Math Exam Resources/Courses/MATH100/December 2016/Question 09 (c) (i)/Solution 1

First, we find the points at which $f''$ is zero or undefined within the domain $\left((-\infty ,-2{\sqrt {2}}]\cup [2{\sqrt {2}},\infty )\right)\setminus \{4\}$ of $f$. We are given that $f''(x)={\frac {8x^{2}(x-3)}{(x^{2}-8)^{3/2}(x-4)^{3}}}$, so the only such points are $x=3$ (where the numerator is zero) and $x=\pm 2{\sqrt {2}},\,4$ (where the denominator is zero).

We use the following sign chart to determine the sign of $f''$.

	$\displaystyle (-\infty ,-2{\sqrt {2}})$	$(2{\sqrt {2}},3)$	$\displaystyle (3,4)$	$(4,\infty )$
$8x^{2}$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$	$+$
$(x-3)$	$\displaystyle -$	$\displaystyle -$	$\displaystyle +$	$+$
$(x^{2}-8)^{3/2}$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$	$+$
$(x-4)^{3}$	$\displaystyle -$	$\displaystyle -$	$-$	$\displaystyle +$
$f''(x)$	$\displaystyle +$	$\displaystyle +$	$\displaystyle -$	$+$
$f(x)$	concave up	concave up	concave down	concave up

This implies that the function is concave down on the interval $\color {blue}(3,4).$