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

Factoring the derivative yields

$f'(x)={\frac {(x-1)(x+1)(x-{\sqrt {6}})(x+{\sqrt {6}})}{(x-{\sqrt {3}})^{2}(x+{\sqrt {3}})^{2}}}$

Now the following sign table is straight forward:

	$(-\infty ;-{\sqrt {6}})$	$(-{\sqrt {6}};-{\sqrt {3}})$	$(-{\sqrt {3}};-1)$	$\displaystyle (-1;1)$	$(1;{\sqrt {3}})$	$({\sqrt {3}};{\sqrt {6}})$	$({\sqrt {6}};\infty )$
$\displaystyle x-1$	$\displaystyle -$	$\displaystyle -$	$\displaystyle -$	$\displaystyle -$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$
$\displaystyle x+1$	$\displaystyle -$	$\displaystyle -$	$\displaystyle -$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$
$x-{\sqrt {6}}$	$\displaystyle -$	$\displaystyle -$	$\displaystyle -$	$\displaystyle -$	$\displaystyle -$	$\displaystyle -$	$\displaystyle +$
$x+{\sqrt {6}}$	$\displaystyle -$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$
$(x-{\sqrt {3}})^{2}$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$
$(x+{\sqrt {3}})^{2}$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$
$\displaystyle f'(x)$	$\displaystyle +$	$\displaystyle -$	$\displaystyle -$	$\displaystyle +$	$\displaystyle -$	$\displaystyle -$	$\displaystyle +$
$\displaystyle f(x)$	$\displaystyle \nearrow$	$\displaystyle \searrow$	$\displaystyle \searrow$	$\displaystyle \nearrow$	$\displaystyle \searrow$	$\displaystyle \searrow$	$\displaystyle \nearrow$

And so we can conclude that the function is increasing on:

$(-\infty ,-{\sqrt {6}})\cup (-1,1)\cup ({\sqrt {6}},\infty )$

and decreasing on:

$(-{\sqrt {6}},-{\sqrt {3}})\cup (-{\sqrt {3}},-1)\cup (1,{\sqrt {3}})\cup ({\sqrt {3}},{\sqrt {6}})$