Finding absolute extrema of function on closed interval

Hello, Jimmy!

This is a very sneaky problem!

Quote:

Given: . $g(x) \:=\: 4\left(1+\frac{1}{x}+\frac{1}{x^2}\right),\;\;[-4,5]$

Find the absolute extrema of the function on this interval.

Set the derivative equal to zero . . .

We have: . $g(x) \;=\;4\left(1 + x^{-1} + x^{-2}\right)$

Then: . $g'(x) \;=\;4(-x^{-2} - 2x^{-3}) \;=\;0\quad\Rightarrow\quad -4\left(\frac{1}{x^2} + \frac{2}{x^3}\right) \:=\:0$

Multiply by $\text{-}\:\!\frac{x^3}{4}\!:\quad x + 2 \:=\:0 \quad\Rightarrow\quad x \:=\:-2$
. . Then: . $g(\text{-}2) \:=\:4\left(1 + \frac{1}{\text{-}2} + \frac{1}{(\text{-}2)^2}\right) \:=\:3$

Critical point: . $(\text{-}2, 3)$

Test the endpoints of the interval.

. . $g(\text{-}4) \:=\:4\left(1 + \frac{1}{\text{-}4} + \frac{1}{(\text{-}4)^2}\right) \:=\:\frac{13}{4} \:=\:3\tfrac{1}{4}$

. . $g(5) \;=\;4\left(1 + \frac{1}{5} + \frac{1}{5^2}\right) \;=\;\frac{124}{25} \;=\;4\tfrac{24}{25}$

I would have written: . $\begin{array}{cc}\text{abs. max.} & \left(5,4\frac{24}{25}\right) \\ \\[-4mm] \text{abs. min.} & (-2,3) \end{array}$

Then I saw that $x = 0$ is a vertical asymptote!

I found that: . $\lim_{x\to0}g(x) \:=\:+\infty\:\text{ from both sides.}$
. . So there is no absolute maximum.

But we do have an absolute minimum: . $(-2,3)$