Question

Use the Close-Interval Method to find the absolute extreme values of the function:

f(x) = |x^(3) - 4x| on [-2,2]

I need the abs max/min values. Also, there is a chart labeled:

x f(x) Extremum
5 empty rows
I think this^ is for the test values, but I'm not sure. Thanks to any help you can give!

Answer 1

So when you're finding absolute maximums and minimums, you need to take the derivative, and then use the point you're given and plug them in. f'(x) = 3x^2 -4...at -2, it would be 12-4 = 8. At 2, it would be the same value, 8.

Answer 2

That is your endpoints, so now use (-2,-1),(-2,0),(-2,1), and so on, using the right value as the plug-in for f(x).

Answer 3

How did you get the other test values (-2,-1),(-2,0),(-2,1)?

Answer 4

& why do you plug in the endpoints to f'(x), and then tell me to plug in the other points into f(x)?

Answer 5

Hmm I can't make much sense of what the other guy was saying...
but umm

Answer 6

So you'll find critical points,
then you'll plug everything, critical points, end points,
into the original function, and compare the outputs.

The largest will be your maximum,
the smallest output will be your minimum.

Do you understand how to find the derivative of this function?\[\large\rm f(x)=|x^3-4x|\]\[\large\rm f'(x)=?\]

Answer 7

isn't it |3x^(2) - 4| ?

Answer 8

If it's absolute value, then it's going to be a little more complicated than that.
Hmm...

Answer 9

does it become a piecewise function?

Answer 10

Let's make a rule for absolute value derivative.
Recall that this is one of the ways that we define absolute value,\[\large\rm |x|=\sqrt{x^2}\]Taking derivative of absolute x gives us:\[\large\rm \frac{d}{dx}\color{orangered}{|x|}=\quad\frac{d}{dx}\color{orangered}{\sqrt{x^2}}=\quad\frac{1}{2\color{orangered}{\sqrt{x^2}}}\cdot (2x)=\quad \frac{x}{\color{orangered}{\sqrt{x^2}}}=\quad \frac{x}{\color{orangered}{|x|}}\]

Answer 11

Yah piecewise would be another way to define it.
Maybe that's better actually, since this function isn't differentiable at x=0...
Hmm.. :d

Answer 12

So anyway, what I was showing is that:\[\large\rm \frac{d}{dx}|\color{orangered}{x}|=\frac{\color{orangered}{x}}{|\color{orangered}{x}|}\]Applying this to our problem,\[\large\rm \frac{d}{dx}|\color{orangered}{x^3-4x}|=\frac{\color{orangered}{x^3-4x}}{|\color{orangered}{x^3-4x}|}\cdot\color{royalblue}{\left(x^3-4x\right)'}\]We also need to chain rule in this problem, ya?
That's what the blue business is about.

Answer 13

\[\large\rm \frac{d}{dx}|x^3-4x|\quad=\quad \frac{(x^3-4x)(3x^2-4)}{|x^3-4x|}\]

Answer 14

oh ok, i understand the derivative

Answer 15

now, i plug the endpoints of [-2,2] into f'(x) right?

Answer 16

no no no :)
We don't care about the slope at the end points.

We want to know the maximum and minimum outputs of the `original function`.

The only place the function can attain a highest or lowest point,
is at the end point or at a critical point.

So all we're doing right now is,
we're looking for critical points.

Answer 17

Here is a simple example of what is going on.
If I have some function on a specific interval,
then it has to reach a maximum height at either the end point, or at one of these critical points (top of a hill)