Science:Math Exam Resources/Courses/MATH100/December 2010/Question 04 (c)

MATH100 December 2010

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Question 04 (c)

Full-Solution Problems. In questions 2-8, justify your answers and show all your work. If a box is provided, write your final answer there. Simplification of answers is not required unless explicitly requested. Let $f(x)={\begin{cases}{\frac {4}{\pi }}\tan ^{-1}x&{\text{if }}x\geq 1\\2-x^{4}&{\text{if }}x<1\end{cases}}$ NOTE: Another notation for tan-1(x) is arctan(x) Determine all critical numbers, open intervals where ƒ is increasing or decreasing, and the x-coordinates of all local maxima or local minima (if any).

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?

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Hint
The function has critical points where ƒ'(x) = 0, or where ƒ'(x) does not exist, but ƒ(x) is defined.

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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. The critical points of a function f(x) are all the values $x$ where 1. $f'(x)=0$ or 2. $f'(x)$ does not exist and $f(x)$ is defined. First, we look at the part of the domain of $f(x)$ where $x>1$. In this domain: $f(x)={\frac {4}{\pi }}\tan ^{-1}(x)\quad \rightarrow \quad f'(x)={\frac {4}{\pi }}\left({\frac {1}{1+x^{2}}}\right).$ We see that there are no values of $x$ such that the derivative is 0 or undefined and hence there are no critical points there. We now move on to the part of the domain where $x<1$: $f(x)=2-x^{4}\quad \rightarrow \quad f'(x)=-4x^{3}.$ Here we see that $f'(x)=0$ for $x=0$. And since $x=0$ is in the part of the domain we are looking at, this is indeed a critical point. Finally, we need to look at $x$ = 1. Remember from part (a) that $f(x)$ is defined and even continuous at $x=1$. To see if the function is differentiable at $x=1$we compute the limit as $x$ approaches 1 from either side. First from the left: $\lim _{x\to 1^{-}}f'(x)=\lim _{x\to 1^{-}}-4x^{3}=-4$ and then from the right: $\lim _{x\to 1^{+}}f'(x)=\lim _{x\to 1^{-}}{\frac {4}{\pi }}\left({\frac {1}{1+x^{2}}}\right)={\frac {4}{\pi }}\cdot {\frac {1}{2}}={\frac {2}{\pi }}$ Since the left and right limits are different we can conclude that the function is not differentiable at $x=1$, which makes that point a critical point. In conclusion, $x=0$ and $x=1$ are the critical points of the function. To determine the intervals of increase and decrease we need to take test points of the function on the intervals, $\displaystyle (-\infty ,0),(0,1),(1,\infty )$ Plugging a value from each interval of $x$ into our expressions for the derivative, we get the following $f'(-1)=4>0,\quad f'(0.5)=-4(0.5)^{3}<0,\quad f'(2)={\frac {4}{5\pi }}>0$ Therefore, $f(x)$ is increasing on the intervals $(-\infty ,0)$ and $(1,+\infty )$ and decreasing on the interval $(0,1)$. By the first derivative test, we can see that the function has a local maximum at $x=0$ and a local minimum at $x=1$. Therefore the function $f$ has a local maximum of 2 (at $x=0$) and a local minimum of 1 (at $x=1$).

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