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 its absolute maximum and absolute minimum values on that interval. This theorem provides the basis for seeking extreme values because it guarantees their existence under the given conditions.

Critical Points

Critical points are points inside the domain of a function where the derivative is either zero or undefined. These points are significant in optimization problems because they are potential locations for local extrema and, by extension, absolute extrema when evaluated along with the endpoints.

Closed Interval Method

The closed interval method is a systematic approach used to find the absolute maximum and minimum values of a function on a closed interval. It involves finding all critical points within the interval, calculating the function’s value at these points as well as at the endpoints, and then comparing these values to identify the highest and lowest.

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the function's rate of change. In optimization problems, differentiation is used to set up and solve for points where the derivative equals zero or does not exist, thereby identifying potential locations for extreme values.

Transcript

00:01 Here we have the function f of x is equal to 1 minus x to the 2 3rd power on the closed interval from negative 8 to 8.

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

00:11 So f prime of x while we take the root of term by term, we have a fractional x point.

00:18 That would be okay, right? so the derivative here is just, well, we just get 2 thirds x to the 2 thirds minus 1.

00:26 That's 2 thirds minus 3 thirds, which is negative 1 third, right? so, and then times negative 1 here in front.

00:35 Therefore, the derivative f prime of x is going to be equal to negative 2 thirds x to the negative 1 3rd.

00:44 Okay, so there's our derivative.

00:47 Now, to find the critical values, we just take our derivative and we set it equal to 0.

00:53 So if we do this, well, this would just imply, right? we could, if we want, i mean, multiply both sides by, what, negative three halves, that's going to cancel out the coefficient on the x, but still make it to zero, right? and then we have, well, x.

01:07 1 3rd equals 0, that's still, that implies that x is equal to 0.

01:13 So this may look a little scary, but it's maybe not, because we just get here that x is equal to 0.

01:20 So there's our critical value.

01:22 Okay, so we list out all critical values and endpoints.

01:25 So we have negative 8 is the one endpoint, then a critical value of 0, and the other.

01:29 End point of 8.

01:31 So now we evaluate our original function at those values.

01:36 So let's do f of 0 first...