Find the absolute extrema of the function on the closed...

Find the absolute extrema of the function on the closed interval. Use a graphing utility to verify your results. (If an answer does not exist, enter DNE.) g(x) = 1/x - 2/x^2, [-2, 1] absolute maximum (x, y) = ( ) (smaller x-value) (x, y) = ( ) (larger x-value) absolute minimum (x, y) = ( )

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Transcript

00:01 In this question, where are asked to find the absolute maximum and the absolute minimum values of the function g over the given interval.

00:07 The only points where the absolute maximum minimum can occur are either the endpoints of the interval or the critical points.

00:20 To find the critical points, we first need to calculate g prime of x.

00:28 G prime of x equals to negative 1 over x squared plus 4 over x cube.

00:50 And we can simplify this to, we're going to write this as negative x plus 4 divide by x cube.

01:05 All right, this is g prime of x.

01:08 Let me rewrite it equals to 4 minus x over x cube.

01:19 Now the critical points are either the points where g prime of x equals to 0, and this happens if x equals to 4, or the points where g prime of x is undefined.

01:39 And this happens if x equals to 0.

01:41 Because in that case we are dividing by 0.

01:45 However, we are not going to consider 0 because our function g is not defined at 0.

02:07 So the only critical point is x equals 4.

02:22 However, note that x equals 4 is not in the interval, from negative 2 to 1, which means that there are no critical points in the interval.

02:50 Therefore, we only need to evaluate our function at the endpoints.

02:56 So, the absolute maximum and minimum value, can only occur at the endpoints...

Question

Find the absolute extrema of the function on the closed interval. Use a graphing utility to verify your results. (If an answer does not exist, enter DNE.) g(x) = 1/x - 2/x^2, [-2, 1] absolute maximum (x, y) = ( ) (smaller x-value) (x, y) = ( ) (larger x-value) absolute minimum (x, y) = ( )

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