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?
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.