The critical points of a function f(x) are all the values
where
1.
or
2.
does not exist and
is defined.
First, we look at the part of the domain of
where
. In this domain:
![{\displaystyle f(x)={\frac {4}{\pi }}\tan ^{-1}(x)\quad \rightarrow \quad f'(x)={\frac {4}{\pi }}\left({\frac {1}{1+x^{2}}}\right).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/226bacca1b6840deb7f1c18238d373735db91b8c)
We see that there are no values of
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
:
![{\displaystyle f(x)=2-x^{4}\quad \rightarrow \quad f'(x)=-4x^{3}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/7171e1c56cdf7ca2957078e9217e4154f710cda4)
Here we see that
for
. And since
is in the part of the domain we are looking at, this is indeed a critical point.
Finally, we need to look at
= 1. Remember from part (a) that
is defined and even continuous at
. To see if the function is differentiable at
we compute the limit as
approaches 1 from either side. First from the left:
![{\displaystyle \lim _{x\to 1^{-}}f'(x)=\lim _{x\to 1^{-}}-4x^{3}=-4}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/8f3878e58b4e43e6fe7778670a9e32e1ad8ed872)
and then from the right:
![{\displaystyle \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 }}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/300edfa5f4a5b5433b78e7289d19cfb727cf0598)
Since the left and right limits are different we can conclude that the function is not differentiable at
, which makes that point a critical point. In conclusion,
and
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 \displaystyle (-\infty ,0),(0,1),(1,\infty )}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d0165f4a0dc3cc8842fa926f57d4cca34b0a7421)
Plugging a value from each interval of
into our expressions for the derivative, we get the following
![{\displaystyle f'(-1)=4>0,\quad f'(0.5)=-4(0.5)^{3}<0,\quad f'(2)={\frac {4}{5\pi }}>0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/7ec8f07d0d43159e5178166a36f31c2e87836f54)
Therefore,
is increasing on the intervals
and
and decreasing on the interval
. By the first derivative test, we can see that the function has a local maximum at
and a local minimum at
.
Therefore the function
has a local maximum of 2 (at
) and a local minimum of 1 (at
).