Find the extrema (that is, the absolute maxima and minima). h(x)=√(x) √((x-3)^2) on [-1,4]

Name: Problem 40, Chapter 4: Calculus
Uploaded: 2021-01-21T16:24:08-08:00
Duration: 4 min 46 s
Channel: Jacquelyn Trost
Description: Find the extrema (that is, the absolute maxima and minima). h(x)=√(x) √((x-3)^2) on [-1,4]

Find the extrema (that is, the absolute maxima and minima)....

Key Concepts

Differentiation

The process of differentiating the function is used to locate critical points. By finding the derivative and setting it to zero or identifying where it is undefined, one can isolate the interior candidates for potential maximum or minimum values on the interval.

Endpoint Analysis

When finding absolute extrema on a closed interval, it is essential to evaluate the function at the endpoints in addition to the critical points. This is because extreme values can occur at the boundaries of the interval, even if the derivative does not indicate a critical point there.

Critical Points

Critical points are points in the interior of the interval where the derivative of the function is either zero or undefined. These points are important because they are potential candidates for local extrema, and hence, for absolute extrema when comparing with function values at the interval’s endpoints.

Extreme Value Theorem

This theorem states that if a function is continuous on a closed interval, it is guaranteed to attain both an absolute maximum and an absolute minimum on that interval. This concept underpins the justification for evaluating potential extrema at both interior points and endpoints of the interval.

Transcript

00:01 For this problem, we have been given the function, h of x equals the cube root of x, and an interval that we're going to be examining, from x going from negative 1 to 8 inclusive.

00:12 Now, the goal of this problem is to find the absolute maximum and minimum values, and then to graph the function and make sure that what we obtained using our calculus matches the graph showing what we expect to see.

00:25 Now, before we get into this problem, it's a good idea for us to stop and check and make sure that we really will have a minimum and maximum value.

00:33 The extreme value theorem says that if you have a continuous function on a closed interval, if both of those are true, then you will indeed have both a minimum and maximum value to your function.

00:49 Well, at a glance, you can see this is indeed a closed interval.

00:53 I'm including both my endpoints, negative 1 and 8, and it is continuous.

00:59 There's no value of x for which i can't find a cube root.

01:05 Remember, since it's an odd root, i can have positives and negatives in there.

01:10 So this is a continuous function.

01:13 So now we're going to find our minimums and maximums.

01:17 Where do we look for those? well, we'll need to examine both endpoints of our interval, as well as any critical points that we obtain from this function.

01:27 Now, as a reminder, critical points occur when my derivative equals 0 or if it's undefined.

01:37 So let's find our first derivative.

01:41 I'm just going to recopy our problem here.

01:43 H of x equals the cube root of x.

01:46 And i'm going to rewrite that.

01:47 Instead of having a root, i'm going to rewrite this as a fractional exponent.

01:51 I find it's easier to take the derivative with exponents versus roots.

01:55 They mean the same thing.

01:57 I just find it much easier.

02:01 First derivative.

02:01 We pull down that exponent and then subtract 1.

02:09 So i have 1 3rd times x to the negative 2 thirds.