# Absolute Extrema

• Mar 11th 2009, 12:29 PM
lmschneider
Absolute Extrema
Locate the absolute extrema of the function f(x)=(2x)/(x^2+1) on the interval [-2, 4]

I don't know where to begin. Do I find the derivative and then plug in -2?
• Mar 11th 2009, 12:40 PM
sinewave85
Quote:

Originally Posted by lmschneider
Locate the absolute extrema of the function f(x)=(2x)/(x^2+1) on the interval [-2, 4]

I don't know where to begin. Do I find the derivative and then plug in -2?

You are on the right track. The first step is indeed to take the derivative. The second is to find those numbers x where f(x) is defined and f'(x) = 0 or f'(x) does not exist. Then you test those critical numbers to see whether they are extrema and if so, what kind.
• Mar 11th 2009, 01:20 PM
lmschneider
So the derivative is confusing me. I have:

[(x^2+1)(2) - (2x)(2x)]/(x^2+1)^2

Then I simplify the top, right?

So:

[(2x^2 +2) - (4x^2)]/(x^2+1)^2

Then, where do I go from there?

Sorry, I'm getting really confused.
• Mar 11th 2009, 01:29 PM
sinewave85
Ok, I have more time than I thought I did, so I can reply more fully.

The derivative describes the slope of the line of the graph of the function, and the maximums and minimums are at those points where the slope is 0. There are no points for which this equation or its derivative is undefined, so we won't go into that.

First you take the derivative of the function:

$\displaystyle f^{\prime}(x) = \frac{(x^2 + 1)(2x)^{\prime} - (2x)(x^2 + 1)^{\prime}}{(x^2 + 1)^2} = \frac{2(x^2 + 1) - 2x(2x)}{(x^2 + 1)} = \frac{2 - 2x^2}{(x^2 + 1)^2}$

Now solve for f'(x) = 0:

$\displaystyle 0 = \frac{2 - 2x^2}{(x^2 + 1)^2}$

$\displaystyle 0 = 2 - 2x^2$

$\displaystyle x = \pm\sqrt{\frac{-2}{-2}} = \pm\sqrt{1} = \pm1$

Both of those critical numbers are in the domain specified, so we can proceed to testing the points using the first-derivative test:

• (x,f(x)) is a relative maximum if
• f'(x) > 0 for all x on an open interval (a,x) to the left of x
• f'(x) < 0 for all x on an open interval (x,b)to the right of x
• (x,f(x)) is a relative minimum if
• f'(x) < 0 for all x on an open interval (a,x)to the left of x
• f'(x) > 0 for all x on an open interval (x,a)to the right of x

Thus we find that at (-1,-1) there is a relative minimum and at (1,1) there is a relative maximum. To prove that those are actually absolute extremums requires the application of the intermediate value theorum, which is a fancy way of saying that by showing that the value of the function at x = -2 is less than 1 and the value of the function at x = 4 greater than -1, you can argue that the function on this interval is never less than -1 or more than 1 and that the points represent absolute extremums.
• Mar 11th 2009, 01:34 PM
sinewave85
Ok, just want to clarify and reinforce that we can apply the intermediate value theorum only because the function is continuous on the closed interval [-2,4].