Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (d)/Solution 1

In order to find the concavity of the function, we first try to find points where the second derivative is zero or undefined. These points will provide endpoints of intervals for concavity.

From part (b) we know that ƒ''(x) is zero exactly when x = 0, and that ƒ''(x) is undefined for x = ±√3

We can now either make a sign table or pick test points between those values to determine whether we are concave up (second derivative positive) or concave down (second derivative negative).

For convenience, recall that

$f''(x)={\frac {2x(x^{2}+9)}{(x^{2}-3)^{3}}}={\frac {2x(x^{2}+9)}{(x+{\sqrt {3}})^{3}(x-{\sqrt {3}})^{3}}}$

	$(-\infty ,-{\sqrt {3}})$	$(-{\sqrt {3}},0)$	$(0,{\sqrt {3}})$	$({\sqrt {3}},\infty )$
$\displaystyle 2x$	$\displaystyle -$	$\displaystyle -$	$\displaystyle +$	$\displaystyle +$
$\displaystyle x^{2}+9$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$
$(x+{\sqrt {3}})^{3}$	$\displaystyle -$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$
$(x-{\sqrt {3}})^{3}$	$\displaystyle -$	$\displaystyle -$	$\displaystyle -$	$\displaystyle +$
$\displaystyle f''(x)$	$\displaystyle -$	$\displaystyle +$	$\displaystyle -$	$\displaystyle +$
Concavity	Concave Down	Concave Up	Concave Down	Concave Up

Therefore $f(x)$ is concave up on $(-{\sqrt {3}},0)\cup ({\sqrt {3}},\infty )$ and concave down on $(-\infty ,-{\sqrt {3}})\cup (0,{\sqrt {3}})$.