Science:Math Exam Resources/Courses/MATH110/April 2017/Question 05 (b)/Solution 1

From (a) we already know that ${\textstyle f'(x)=-10x(4-x^{2})^{4}}$, then by product rule we get the second derivative that

${\begin{aligned}f''(x)&=(f'(x))'=(-10x(4-x^{2})^{4})'\\&=(-10x)'(4-x^{2})^{4}+(-10x)[(4-x^{2})^{4}]'\\&=-10(4-x^{2})^{4}+80x^{2}(4-x^{2})^{3}=(4-x^{2})^{3}\cdot (90x^{2}-40)=10(2-x)^{3}(2+x)^{3}(3x+2)(3x-2).\end{aligned}}$

The last equality follows from $(4-x^{2})=(2-x)(2+x)$ and $(90x^{2}-40)=10(9x^{2}-4)=10(3x+2)(3x-2).$

Then, we can see that the zeros of ${\textstyle f''}$ are ${\textstyle -2,-{\frac {2}{3}},{\frac {2}{3}},2}$. At these points, $f''$ possibly changes its sign. Therefore, based on these zeros, we make a partition of intervals on real line and figure out in which interval $f''$ has positive or negative signs. First, we consider the sign of each factor and then combine them to get the sign of $f''$. Note that $(2-x)$ has the same sign with $(2-x)^{3}$, and similarly $(2+x)$ has the same sign with $(2+x)^{3}$.

Sign of $f''$

	$(-\infty ,-2)$	$(-2,-3/2)$	$(-3/2,3/2)$	$(3/2,2)$	$(2,+\infty )$
$(2-x)^{3}$	$+$	$+$	$+$	$+$	$-$
$(2+x)^{3}$	$-$	$+$	$+$	$+$	$+$
$3x+2$	$-$	$-$	$+$	$+$	$+$
$3x-2$	$-$	$-$	$-$	$+$	$+$
$f''$	$-$	$+$	$-$	$+$	$-$

Recall that ${\textstyle f}$ is concave up (convex) on the interval where ${\textstyle f''>0}$; ${\textstyle f}$ is concave down on the interval where ${\textstyle f''<0}$. Thus ${\textstyle f}$ is concave up on ${\textstyle \color {blue}(-2,-{\frac {2}{3}})\cup ({\frac {2}{3}},2)}$ and concave down on ${\textstyle \color {blue}(-\infty ,-2)\cup (-{\frac {2}{3}},{\frac {2}{3}})\cup (2,\infty ).}$