Find the absolute maximum and minimum values of each function over the indicated interval, and i

Find the absolute maximum and minimum values of each...

Key Concepts

Endpoint Evaluation

When dealing with functions on a closed interval, it is necessary to evaluate the function at the endpoints of the interval in addition to any critical points. The absolute maximum and minimum values can occur either at endpoints or at the identified critical points.

Absolute Maximum and Minimum

The absolute maximum is the highest function value achieved on the interval, and the absolute minimum is the lowest. Determining these values involves comparing the function outputs at all critical points and the endpoints, selecting the largest and smallest values respectively.

Extreme Value Theorem

This theorem states that if a function is continuous on a closed interval, then it must attain both an absolute maximum and an absolute minimum value on that interval. This is the foundational concept that guarantees solutions exist for optimization problems on closed intervals.

Critical Points

Critical points are values in the domain where the first derivative of the function is zero or undefined. In the context of finding extreme values, these points are important because they are candidates where local maxima or minima may occur.

Transcript

00:01 Here we consider the function f of x is equal to x cubed minus x to the fourth over the closed interval from negative 1 to 1.

00:08 So again, we first take the derivative of our function.

00:12 So we have f prime of x.

00:15 I would just take derivative here turn by term and see that the derivative is just 3x squared minus 4x cubed.

00:25 Okay, so there's a derivative.

00:27 Now the critical values are going to occur where, the derivative is equal to 0.

00:32 So we find the values of x that make the derivative 0.

00:35 So we take our derivative, we set it equal to 0, and then solve for x.

00:39 Notice we can factor out and x squared here.

00:42 We do that.

00:43 We have x squared times 3 minus 4x.

00:50 Okay, and that's still that equal to 0.

00:52 So we see that either x squared is equal to 0.

00:55 That implies that x is equal to 0, or 3 minus 4 x, equal to 0, that implies that negative 4x is equal to negative 3, which implies that x is equal to three -fourths.

01:09 So we have that x is equal to 3 .4s would make the derivative equal to 0.

01:16 So those would be our critical values.

01:20 Okay.

01:21 And then, well, notice, i mean, then this list out the critical values along with the endpoints, because the absolute max and min has to occur at either one of the endpoints or at a critical value where the derivative is equal to zero.

01:36 Okay, so we list out our endpoints and critical values...

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