From (a) we already know that
, then by product rule we get the second derivative that
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/a1c80aaaacadc1fc08c409282c07a36e9c64b0c1)
The last equality follows from
and
Then, we can see that the zeros of
are
. At these points,
possibly changes its sign. Therefore, based on these zeros, we make a partition of intervals on real line and figure out in which interval
has positive or negative signs. First, we consider the sign of each factor and then combine them to get the sign of
. Note that
has the same sign with
, and similarly
has the same sign with
.
Sign of
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Recall that
is concave up (convex) on the interval where
;
is concave down on the interval where
. Thus
is concave up on
and concave down on