1
$\begingroup$

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):

enter image description here

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!

$\endgroup$
5
  • 2
    $\begingroup$ Equating the first derivative to zero, we get 2x-1=0. Therefore, x = 1/2. If i'm not wrong, that's the only critical point. And you can test the sign (positive/negative) of the first derivative around x = 1/2. If the derivative goes from positive on the left of 0.5 to negative to the right of 0.5, 0.5 is a maxima. Otherwise, it is a minima. $\endgroup$
    – NoLand'sMan
    Commented May 15, 2019 at 15:11
  • $\begingroup$ Yeah that makes sense, but if you plug x=1/2 into the original function, you get $i*tanh(\pi -ilog(5/4))$ (per Wolfram). how would a complex value be a min/max? I'm honestly convinced there might be a typo in this problem since it was meant for Calc 1 students $\endgroup$
    – Anthony
    Commented May 15, 2019 at 15:15
  • 1
    $\begingroup$ How did you get $i$? (1/2)^2-1/2-1 = -5/4. Taking the absolute value gives arctan(log(5/4)) right? $\endgroup$
    – NoLand'sMan
    Commented May 15, 2019 at 15:22
  • $\begingroup$ darn it. forgot the absolute value when plugging into the function lol. Thanks! $\endgroup$
    – Anthony
    Commented May 15, 2019 at 15:23
  • 1
    $\begingroup$ Glad I could help! $\endgroup$
    – NoLand'sMan
    Commented May 15, 2019 at 15:25

1 Answer 1

Reset to default
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .