Science:Math Exam Resources/Courses/MATH110/April 2017/Question 05 (a)

MATH110 April 2017

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Question 05 (a)
Let $f(x)=(4-x^{2})^{5}$. (a) Find the intervals where $f$ is increasing and the intervals where it is decreasing.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

In determining intervals where a function is increasing or decreasing, you first find domain values where all critical points will occur; then, test all intervals in the domain of the function to the left and to the right of these values to determine if the derivative is positive or negative. If ${\textstyle f'(x)>0}$, then f is increasing on the interval, and if ${\textstyle f'(x)<0}$, then f is decreasing on the interval

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We need first to find the derivative of ${\textstyle f}$, which is by chain rule $$f'(x)=5(4-x^{2})^{5-1}(4-x^{2})'=5(4-x^{2})^{5-1}(-2x)=-10x(4-x^{2})^{4}.$$ The critical points are the points where ${\textstyle f}$ is zero, that is by factoring $$-10x(4-x^{2})^{4}=-10x(2-x)^{4}(2+x)^{4}=0.$$ So critical points are ${\textstyle 0,2,-2}$ (multipilicty). It is not hard to see that ${\textstyle f'>0}$ when ${\textstyle x<-2}$; ${\textstyle f'>0}$ when ${\textstyle -2<x<0}$; ${\textstyle f'<0}$ when ${\textstyle 0<x<2}$ and ${\textstyle f'<0}$ when ${\textstyle x>2}$. (Remark: since ${\textstyle (2-x)^{4}(2+x)^{4}}$ is always nonnegative, so the only part determining positivity and negativity is the factor ${\textstyle -10x}$). Recall that If ${\textstyle f'(x)>0}$, then f is increasing on the interval, and if ${\textstyle f'(x)<0}$, then f is decreasing on the interval. In all ${\textstyle f}$ is increasing on ${\textstyle \color {blue}(-\infty ,-2)\cup (-2,0)}$ and decreasing on ${\textstyle \color {blue}(0,2)\cup (2,\infty )}$.