Science:Math Exam Resources/Courses/MATH110/April 2018/Question 03 (c)/Solution 1

The first step is to find ${\textstyle f''(x)}$. We do this using the chain rule. Write ${\textstyle f'(x)=g(h(x))}$, where ${\textstyle g(u)=u^{4}}$ and ${\textstyle h(x)=x^{2}-9}$. Then the chain rule with $g'(u)=4u^{3}$ and $h'(x)=2x$ gives

$$f''(x)=(f'(x))'=g'(h(x))\cdot h'(x)=4(x^{2}-9)^{3}\cdot 2x=8x(x^{2}-9)^{3}.$$

Next, we find the roots of ${\textstyle f''(x)}$ by setting ${\textstyle f''(x)=0}$ and solving for ${\textstyle x}$. If ${\textstyle 8x(x^{2}-9)^{3}=0}$, then either ${\textstyle x=0}$ or ${\textstyle (x^{2}-9)^{3}=0}$. In the latter case, we obtain two solutions: ${\textstyle x=3}$ and ${\textstyle x=-3}$.

We have found three possible inflection points. Note that we are not quite done yet: not every point at which the second derivative is zero is necessarily an inflection point. For each of the three solutions we found, we need to check whether the second derivative of ${\textstyle f}$ changes sign at these points. Since the solutions are isolated, it suffices to check the sign of the second derivative at one point on either side of each of the solutions.

We have

$$f''(-4)<0;$$

$$f''(-1)>0;$$

$$f''(1)<0;$$

$$f''(4)>0.$$

This shows that each of the three points ${\textstyle x=-3,0,3}$ are indeed inflection points.

Answer: The inflection points are ${\textstyle {\color {blue}x=-3,0,3}}$.