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Classify all critical points.$h(x)=\left(12-x^{2}\right)^{1 / 2}$

$$\mathrm{m}(0,2 \sqrt{3})$$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 2

The First Derivative Test

Derivatives

Harvey Mudd College

Baylor University

University of Nottingham

Idaho State University

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

01:55

Locate all critical points…

05:05

Classify all critical poin…

02:09

03:53

01:37

04:05

02:33

01:50

Determine all critical poi…

02:00

03:24

Locate and classify all th…

01:59

02:55

I want to find the derivative of this function. We do that by using chain rules. We get 1/2 12 minus X squared The native 1/2. So that's our derivative of the outer function times the derivative of the inter function which is native to X. We want this to equal to zero. Again we got the -2 x. by bringing this down and 12% is a constant. So so to find out if equals zero, we have a product of two expressions. So when each of those expressions equal zero, the whole equation will equal zero. So for the first one, when X equals zero is actually when Y equals zero or they drifted will equal zero. And so here for the inner function we can set that equal zero. That's what we did up here was actually setting physical this many expression to zero and solving for X. So we can do the same here we get Plus reminds us we're out of 12 because keep in mind Yeah, square to both sides. And then we take the square root of 12 to get X. And so this since we're taking the square root of something, it can either be positive or negative. So next you want to plug each of these values into our original equation to get the corresponding Y values. And when you do that you get y equals zero for both positive and negative square root of 12. And we get Y is equal to the square root of 12 here, the positive one. Next we're going to plot these values on a number line and pick values slightly above and slightly below them. So and compare that to figure out if the critical points are increasing or decreasing and what makes that? So we got zero here Square root of 12. A Negative Square Root of 12. So we know at these points The Derivative Equals zero. So we're gonna pay values above and below. Let's pick eight of 1, one and Let's do five and -5. So a fiber negative five. We actually get negative values for our original expression here. And keep in mind when something is raised to the one half. That actually means the square root of something and we cannot have rationally you can't have square root of negative values. So here and here the function is undefined. So at native one and one if we were to plug that into our original equation, we get increasing and decreasing, which makes that a local maximum.

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