Absolute maxima and minima Determine the location and value of the absolute extreme values of f

Absolute maxima and minima Determine the location and value...

Key Concepts

Absolute Extrema

Absolute extrema, also known as global extreme values, represent the maximum and minimum values that a function attains over a particular interval. In the context of continuous functions on closed intervals, these values can occur either at critical points found within the interval or at the endpoints of the interval.

Extreme Value Theorem

The Extreme Value Theorem guarantees that a continuous function on a closed interval has both an absolute maximum and an absolute minimum. This theorem is fundamental in calculus, as it underpins the method of evaluating a function's values at critical points and at the endpoints to determine its absolute extreme values.

Critical Points

Critical points are points in the domain of a function where its derivative is either zero or undefined. These points are candidates for local extrema, which can also be the absolute maximum or minimum when evaluated over a closed interval. Identifying critical points is a key step in the process of finding a function’s extreme values.

Differentiation

Differentiation is the process of computing the derivative of a function, which provides information about the function's rate of change. The derivative is used to find the critical points by setting it equal to zero. Analyzing these points helps determine where the function might achieve local and, potentially, absolute extrema.

Endpoint Evaluation

When finding absolute extrema on a closed interval, the function must be evaluated not only at critical points but also at the endpoints of the interval. This is essential because the absolute maximum or minimum might occur at an endpoint rather than at a critical point within the interval.

Transcript

00:01 Let's spend the absolute maximum and minimum values for this function.

00:05 This 4x to the third over 3 plus 5x squared minus 6x on this interval from negative 4 to 1.

00:12 So the first thing we need to do is find the critical values or the function or the derivative of the function is equal to 0.

00:19 So let's find the derivative.

00:21 So f -parl of x.

00:23 This is a polynomial.

00:24 So it's pretty nice.

00:25 So this becomes 4x squared.

00:28 And multiply 3 down.

00:29 It cancels.

00:30 Reduce the power by 1 plus 10 x minus 6.

00:37 Okay, now we need to figure out where this is equal to 0.

00:47 So the first thing i'm going to do is i'm going to factor out 2, so this becomes 2x squared plus 5x minus 3.

01:00 This is equal to 0.

01:01 And now you could have done, you could have done, you could have have just done the quadratic formula for this one right away, but i'm going to do it for this part here.

01:11 I think it's a little easier.

01:13 So we're going to have negative 5 plus or minus square root of 25.

01:21 That's 5 squared, minus 4 times 2 times negative 3 all over 2 times 2 on the bottom, so that's 4.

01:32 Let's see, that's equal to...

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