I'm trying to find the local/global extrema of $\arctan(\log|x^2-x-1|)$ where $-2\leq x \leq 4$.
So what I did was take the first derivative (computed below):
Then to find the critical points, Wolfram Alpha says it's $x=1/2$
But how would I determine which values are global/local extrema from here? The only thing I can think of is plugging in $x=1/2$ and the lower/upper bound $(-2,4)$.
I can't possibly imagine taking the second derivative of that function because it'd be ridiculous.
I'm trying to keep things simple because this is for a Calc 1 class I'm TA'ing, so the students definitely aren't expected to take the second derivative of that, but I'm struggling to solve the problem in general.
Any help would be very much appreciated!
The domain of $$f(x)=\tan^{-1}[\log(x^2-x-1)]$$ is $x < \frac{1-\sqrt{5}}{2}$ or $x>\frac{1+\sqrt{5}}{2}$. But the assumed domain is $[-2,4]$. Finally the function is to be considered in the domains $D_1=[-2, \frac{1-\sqrt{5}}{2})$ and $D_2=(\frac{1+\sqrt{5}}{2}, 4].$ As the derivative $$f'(x)=[1+\log^2(x^2-x-1)]^{-1} (x^2-x-1)^{-1} (2x-1)$$ is negative in $D_1$ so $f(x)$ is decreasing and in $D_2$ it is positive so $f(x)$ is increasing in $D_2$. So the global maximum is either $f(-2)$ or $f(4)$. The former is 1.01 and the latter is 1.17. Hence, the global maximum is $f(4.)$ and there is no global minimum.