Find the absolute maximum and minimum of $f(x) = \frac{1}{x} - \frac{2}{x^2}$ on $[-2,1]$ Note: Turns out this is a pretty tricky problem where we need to use techniques not developed in the chapter. We will go over this problem in class on Tuesday.
Find the absolute maximum and minimum of $f(x) = \frac{1}{x} - \frac{2}{x^2}$ on $[-2,1]$

To find a the extrema of a continuous function $f$ on a closed interval $[a,b]$, use the following steps:
1) Find the critical numbers of $f$ on the open interval $(a,b)$.
2) Evaluate $f$ at each of its critical numbers in $(a,b)$
3) Evaluate $f$ at the endpoints, $a$ and $b$
4) The least of these values is the absolute minimum, the greatest of these is the maximum.

Now, unfortunately, our function is not defined when $x=0$ and so our function is not continuous on $[-2,1]$, which means there might be more to finding the maximum and minimum then simply following the above steps. Lets see what goes wrong and how we can still solve the problem.
$Solution:$
First lets gather all the points where an absolute extrema could possibly happen. These are the values of $x$ where $f'(x)=0$ and the boundary points of our domain $[-2,1]$ (so $x=-2$ and $x=1$). Let's find the other values $x$ where an absolute extrema could happen by solving $f'(x)=0$:
$f(x) = \frac{1}{x} - \frac{2}{x^2} = x^{-1} - 2x^{-2}$ 
$f'(x) = -1x^{-2}-2(-2x^{-3}) = -x^{-2}+4x^{-3}$
Now we have to solve $-x^{-2}+4x^{-3}=0$.
$\rightarrow$
$4x^{-3}=x^{-2}$. 
Let's divide both sides by $x^{-2}$ ( The same thing as multiplying both sides by $x^{2}$)
$4x^{-1}=1$
Now lets multiply both sides by $x$:
$4=x$
Cool, so the only solution to $f'(x)=0$ is $x=4$... But $x=4$ is not in our domain of $[-2,1]$, so it's not relevant.
So the only two values of $x$ where an absolute extrema could occur is $x=-2$ and $x=1$. 
Now lets plug in the possible $x$ values of the absolute extrema into $f(x)$:
$f(-2) = \frac{1}{-2} - \frac{2}{(-2)^2} = -\frac{1}{2} - \frac{2}{4} = -1$
$f(1) = \frac{1}{1} - \frac{2}{1^2}=-1$
Since there are only two possible values of $x$ that could be absolute extrema, and they both have the same $y$ value when you plug them into our function, it looks like we can't have both an absolute maximum and absolute minimum (well, perhaps if the function was constant like $f(x)=5$ then 5 would be the absolute maximum and the absolute minimum). 
So we have to figure out whether $(-2,1)$ and $(1,-1)$ are absolute maximum, absolute minimums, or neither. We can do a similar process we do when we are looking for relative maximum and minimum, where we make a number line to see where the function is increasing and decreasing. When we do this, we have to include $x=0$ as a critical point since $f'(x)$ is not defined there.
I want to remind you all that usually we do not have to do this when finding absolute extrema, and we only do because our function is not defined everywhere in our domain.
So if we make a number line and seperate it with critical points $x=-2$, $x=0$, $x=1$, and $x=4$ then plug in numbers from each section (well we can skip $(4,\infty)$, since it's outside the domain that we care about):
$f'(3) = \frac{-3}{27}+\frac{4}{27}>0$
$f'(-1) = -5 < 0$
$f'(.5) = -(-.5)^{-2}+4(-.5)^{-3}=-4+32=28>0$
So our function is increasing as $x$ increases towards $-2$, then our function is decreasing as $x$ increases towards $0$, then our function again begins increasing as $x$ increases towards $1$. This is all the information we need to see that $(-2,-1)$ and $(1,-1)$ must be absolute maximums.
This is because when you write this out on the number line, and remember that we are only interested in values of $f(x)$ on the domain $[-2,1]$, you can see that the maximum value of the function will either happen at $x=-2$ or $x=1$, and since we know $f(-2)=f(1)$, we have both $(-2,-1)$ and $(1,-1)$ must be absolute maximums.
Furthermore, since we have that $x=-2$ and $x=1$ are the only points in our domain that could be absolute extrema, and they both correspond to absolute maximums, we have that the absolute minimums DNE.
 A: A critical point is an $x$ such that $f'(x) = 0$ or $f'(x)$ is undefined.  (Many people forget the second case.)  (From the Wikipedia: "A critical point of a function of a single real variable, $f(x)$, is a value $x_0$ in the domain of $f$ where it is not differentiable or its derivative is $0$.")
(Aside:  I've long held that we should include endpoints of the interval as critical points, for the same reason that we include points where the derivative is undefined: the derivative is incapable of indicating that such a point is an extremum.  But this hasn't caught on.)
Since $f'(x) = -x^{-2} + 4x^{-3} = (4-x)x^{-3}$, $f'$ is undefined when $x = 0$ and $f'(x) = 0$ when either $4-x = 0$, so $x = 4$, or when $x^{-3} = 0$, so never.  Thus, the list of potential critical points is $\{0,4\}$.  Of these, only $0$ is in the interval $[-2,1]$.  Then $\lim_{x \rightarrow 0} f(x) = -\infty$ and we discover there is no absolute minimum.
To find the absolute maximum (or verify it does not exist), we need to check the other critical points in the interval (of which there are none) and endpoints.  $f(-2) = -1$ and $f(1) = -1$, so the absolute maximum on this interval is $-1$, attained at $x = -2$ and at $x = 1$.  (Recall that maxima and minima are values, that is, outputs of the function, not points on the graph.)
Letting a computer graph this for us,

we see that the behaviour is as described.
