Science:Math Exam Resources/Courses/MATH100/December 2010/Question 04 (c)/Solution 1

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 $x=1$).