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

Extreme Value Theorem

The Extreme Value Theorem states that if a function is continuous on a closed interval, then it must attain both an absolute maximum and an absolute minimum within that interval. This theorem is fundamental because it guarantees the existence of extrema, making optimization problems on closed intervals well-defined.

Critical Points

Critical points are locations in the domain of a function where its derivative is zero or undefined. These points are important because they are candidates for local minima and maxima, and in the context of bounded intervals, they must be examined alongside endpoints to determine the absolute extrema.

Differentiation

Differentiation is a key tool in calculus used to compute the derivative of a function. The derivative provides information about the function’s rate of change and helps identify critical points. This is essential for optimization problems, as setting the derivative equal to zero can reveal potential extrema.

Closed Interval Method

The closed interval method involves finding the absolute maximum and minimum values of a continuous function on a closed interval by evaluating the function at all critical points and at the endpoints of the interval. This process ensures that all possible extrema are considered, in accordance with the Extreme Value Theorem.

Transcript

00:01 Here is f of x is equal to x to the fourth minus 8x squared plus 3 over the closed interval from negative 3 to 3.

00:08 So first we're going to take the derivative of our function.

00:11 So our function is f of x, so therefore our derivative is f prime of x.

00:15 So we just differentiate here turn by term.

00:18 Can we get that our derivative is 4x cube minus 16 x to the 1st, so minus 16 x.

00:26 Okay, and then we find our critical values.

00:31 So our critical values occur where the derivative is equal to zero.

00:34 We take a derivative and we set it equal to zero and then solve for x.

00:38 But notice that we can definitely factor out here a factor of what, of 4x, right? so we factor out of 4x.

00:47 So we have 4x times x squared minus 4, right? and that's equal to zero.

00:58 Okay.

00:59 Oh, but this is going to be a factor down further, right? so x minus four, that factors.

01:04 We have four x times.

01:05 Well, this is going to be what, x plus two times x minus two, right? okay, so this is then still equal to zero.

01:14 So now we see that either for x is equal to zero, which implies that x is equal to zero or x plus two is equal to zero, which implies that x would be equal to negative two.

01:25 Or x minus 2 is equal to 0, which implies that x would be equal to positive 2.

01:31 So we then list out all of our critical values and our endpoints, because the absolute max or min would have to occur where the derivative is equal to 0 at one of the critical values or at one of the endpoints.

01:45 So our critical values and endpoints here would be negative 3, negative 2, 0, 2, and 2, and 3, and then we go back and we evaluate the original function, f of x, at these values.

02:03 So let's first evaluate f of zero.

02:06 So we have f of zero...

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