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

Factoring the derivative yields

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

Now the following sign table is straight forward:

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

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

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

and decreasing on:

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