Locating critical points Find the critical points of the follo...

Question

Locating critical points Find the critical points of the following functions. Assume a is a nonzero constant.

ƒ(x) = x √(x-a)

Accepted Answer

Welcome back, everyone. In this problem, we want to determine the X coordinate of the critical point of the function F X equals 2 X multiplied by the square root of X minus 2B, where B is a constant and B is greater than 0. A says X equals 4B. B says it's half of B, C says it's B, and D says it's 2B. Now what do we know that's going to help us to determine the X coordinate of the critical point of our function? First, let's ask ourselves, what do we know about critical points? Well, recall, OK. That the critical point. The critical point. Of a function And it's their function FFX. Is any point, OK, any point X where X equals C in the domain, OK, that satisfies either of the following two conditions. Where first either the first derivative. Of F of C is equal to 0, OK. Or second or first derivative of FOC is undefined, OK? Of course it's undefined where F of X is defined. So these are the two conditions under which we'll be able to find the critical point, OK, of FFX. So notice that in both of these conditions, we'll need the first derivative of FFX. So let's go ahead and find the first derivative and then try to find the critical points where either the first derivative is zero or where it is undefined. So to do that, we'll need to differentiate FF X. Now, here, notice that FF X is a product and to differentiate a product, we can use the product rule of differentiation. Recall the product rule basically tells us that if we have a product Y that's equal to UV where you and V are. Uh, functions of X. Then the derivative of Y equals U V plus UV where U and V are the derivatives of U and V respectively. So in this case, if we let you be equal to 2 X. And we let v be equal to the square root of X minus 2 b. Then the derivative of u will be equal to 2, while the derivative of V will be equal to 1 divided by 2 multiplied by the square root of X minus 2 B. eh. I know In that case, we can go ahead and differentiate FFX because that means our derivative. Is gonna be equal to UV. OK, so that's 2 multiplied by the square of X minus 2 B. Plus, you multiplied by V. So that's 2 X multiplied by 1 divided by 2 root X minus 2B. Now, we can combine these into one fraction, OK, which is gonna be 2 multiplied by X minus 2B plus X, all divided by the square root of X minus 2B. And we can expand this further in our numerator, OK, cause 2 multiplied by X is 2 X, 2 multiplied by negative 2B is -4B, and 2 X. Plus X equals 3 X, then we're gonna have minus 4B and all of this is gonna be divided by the square root of X minus 2B. I know what's important here to note is that our function exists, OK, where Our derivative, sorry, or derivative, I should have written that here, our derivative exists where X minus 2B is greater than 0, which means then that X must be greater than 2B. Know that we have our first derivative, then we can test both of our conditions. Let's start with the first one. When our first derivative. Is equal to 0, OK. Then will we find a critical point? Can we solve for X? Well, in that case, that means 3 X minus 4B. Divided by the square root of X minus 2b will be equal to 0. And if we multiply both sides of our equation by this root X minus 2B, then that means 3X minus 4B will be equal to 0, which makes X equal to 4B divided by 3. Now, is this value a possible value of X? Well, if we think about it here, remember earlier we said X is greater than 2B and 4/3 of B is not greater than 2B, OK? So since 4/3 of B, OK, is less than 2B, it's not greater than 2B, then that means this is not a value of X and as such, there is no critical point, OK? There is no critical point. When the first derivative is equal to 0. Now let's take the second condition. What about when our first derivative is undefined? Well, let's think about it. For our quotient, for our first derivative, what will make it undefined? Well, we know that it will be undefined if the denominator is equal to 0. So in that case, root X minus 2b will be equal to 0. And if we square both sides, then X minus 2B will be equal to 0, and thus X is going to be equal to 2B to make our denominator 0 to make our function undefined. Now, is this a plausible value of X? OK. In other words, does it work? Is our original function defined here? Well, at X equals to b. Then our function f of 2B is going to be equal to 2 multiplied by 2B multiplied by the skirt of 2 b minus 2B, and 2 b minus 2B is 0.0 multiplied by everything here equals 0. That means our function is defined. And since it's defined, therefore, that means that the X coordinate. OK, X coordinate. Of the critical point. Is X equals to B. If we look back at our answer choices, that means D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Locating critical points Find the critical points of the following functions. Assume a is a nonzero constant.

ƒ(x) = x √(x-a)

Key Concepts

Critical Points

Derivative

Function Behavior

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