Checking the Mean Value Theorem

Question

Find the value or values of c that satisfy the equation (f(b) − f(a)) / (b − a) = f′(c) in the conclusion of the Mean Value Theorem for the functions and intervals in Exercises 1–6.

f(x) =√(x − 1), [1, 3]

Accepted Answer

Welcome back, everyone. For the function F of X equals the square root of X + 4 and the interval -30, find the value or values of C that satisfy the equation F of B minus F of A divided by B minus A equals F C according to the mean value theorem. A says C is -1, B-2, C -1.75, and the D 1.75. Now, in essence, what we're trying to figure out here is a value of C that makes this equation true. Now, first of all, we have to ask ourselves what is B, what is A, then we need to find F of B and F of A, then the derivative of F of X to find F of C. Now, given the interval, OK? Given our interval. Of -3 to 0, OK, then that tells us A would be equal to -3. And B will be equal to 0. If that's the case, that means that FF B is going to be FF 0. And if we put 0 into our function, then this is gonna be the square root of 0 + 4, which is root 4, which equals 2. Next, FFA is gonna be F of -3. And the square root of -3 + 4 is gonna be the square root of 1, which equals 1. So F of B is 2 and FFA is 1. Next, we need to find F C. Now to find that, that means we're going to differentiate FFX first, OK? So since, let me write that here, since FF X. Is equal to the square root of X +4, which equals X + 4 arrays to the power of a half. OK, we could rewrite it this way using what we know about powers. Then to find the derivative of F of X, we can use the chain rule of differentiation since it's a composite function, which means we'll differentiate the outer function. So we bring down the half and then subtract one from the function. So a 1/2 minus 1 is negative a half. And then we multiply that by the derivative of the inner function and the derivative of X + 4 is 1. Thus, the derivative of F of X is going to be equal to 1 divided by 2 multiplied by root X plus 4. OK? That's how we could write it. And since this is the derivative of F of X, then this implies that F of C, we're gonna substitute uh C or we're gonna we're gonna put C wherever we see X. So F of C is gonna be one divided by 2 root C plus 4. Now that we have F C, F of B, and F of A, we can substitute them into our equation to see what value of C will make it true. In other words, we can now solve for C according to the mean value theorem. So no, this would imply then that we're gonna have F of 0, which is 2. Minus F of -3, which is 1 divided by 0 minus -3. Equal to 1 divided by 2 multiplied by the square root of C + 4. No, 2 minus 1 is 10 minus -3 is 0 + 3, which is 3. So 1/3 equals 1 divided by 2 root C + 4. Now if that's the case. And if we cross multiply here, then 2 root C + 4 is going to be equal to 3, OK? Which means that the square root of C + 4 equals 3 divided by 2 or 3 halves. If I come over to the right here, OK, then this means that if we square both sides, C + 4 is going to be equal to 3 halves squared, which is 9/4, and thus C is going to be equal to 9/4 minus 4. Which, if we calculate here, that's really 9/4 minus 16/4 and that's gonna be negative 7/4. As a decimal number, 7/4 is -1.75. Thus, the value of C that will make the equation true is going to be -1.75. And we have to first, we also have to make sure that this value falls into the interval for it to be true. And the C equals -1.75 is in the open interval from -3 to 0. So it satisfies. Uh let's say here since C. Is in. The the open interval. Yeah It satisfies. The condition Of the mean value theorem. OK? So we're sure that we are correct. C equals -1.75. When we look back on our answer choices, that means C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Checking the Mean Value Theorem

Find the value or values of c that satisfy the equation (f(b) − f(a)) / (b − a) = f′(c) in the conclusion of the Mean Value Theorem for the functions and intervals in Exercises 1–6.

f(x) =√(x − 1), [1, 3]

Key Concepts

Mean Value Theorem

Continuity and Differentiability

Derivative Calculation

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