Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (b)/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.

$2x(x^{2}+9)=0.$

Since x²+9=0 has no real solutions, the only value of x that satisfies the above equation is when x=0.

To determine where f"(x) does not exist, we set the denominator to 0, i.e.

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

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) as $x\rightarrow a$ to determine if f"(x) did not exist or was zero (or another finite number).