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

To determine where f'(x)=0, we look at the numerator of f'(x) and note that f'(x)=0 when the numerator is zero, i.e.

$(x^{2}-1)(x^{2}-6)=0,$

which implies x²-1=0 or x²-6=0. Solving each equation and noting that both positive and negative solutions are valid, we obtain

$f'(x)=0{\text{ if and only if }}x=\pm 1{\text{ or }}x=\pm {\sqrt {6}}.$

To determine where f'(x) does not exist, we set the denominator to 0:

$x^{2}-3=0.$

So, f'(x) does not exist when $x=\pm {\sqrt {3}}$.

Remark: Note that the denominator and numerator are never simultaneously 0. If they were, say at x=a, we would have to take the limit of f'(x) for $x\rightarrow a$ to determine if f'(x) did not exist or was zero (or another finite number).