Checking the Mean Value Theorem

Question

Which of the functions in Exercises 7–12 satisfy the hypotheses of the Mean Value Theorem on the given interval, and which do not? Give reasons for your answers.

f(x) = √(x(1 − x)), [0, 1]

Accepted Answer

Welcome back, everyone. In this problem, we want to evaluate if the function F of X equals the square root of X multiplied by 5 minus X on the interval 05 meets the criteria of the mean value theorem. A says the function satisfies the hypothesis of the mean value theorem, and B says the function does not satisfy the hypothesis of the mean value theorem. Now what criteria does our function need to meet when it says it needs to meet it for the mean value theorem? Where recall That if a function. Is continuous. On a closed interval. And differentiable. On an open interval and we're talking about the same interval. OK. Then that means. It satisfies the criteria of the mean value theorem. So now what that means for us is that we need to test if our function is continuous on our closed interval, in this case from 0 to 5, and then if it is differentiable on the open interval for 0 to 5. So let's try to do that. First, let's test its continuity, OK, on the closed interval from 0 to 5. So, is it continuous? That's what we're trying to figure out. Well, if we take our function, OK, F of X is equal to the square root of X multiplied by 5 minus X, then we can tell here that it involves a square root. So that means this expression under the root has to be positive for our function to be real. So we need to check if this expression is non-negative on this interval. Now if we think about it here, if that's true, then 4 x multiplied by 5 minus X, it must be greater than or equal to 0. And for this to be true, then either X has to be greater than or equal to 0 or 5 minus X has to be greater than or equal to 0, OK? And if X is greater than or equal to 0. Then if we look at our interval that goes from 0 to 5, we know that this condition is satisfied, OK? So we know that this will always be true. If 5 minus X is greater than or equal to 0, that means X must be less than or equal to 5. And again, when we look at our interval, it goes from 0 to 5. So that means that this condition is also satisfied. So since both of these conditions are satisfied, we know the expression under the root will always be a non-negative number and thus. That means FFX is continuous on this closed interval, OK? So we've satisfied the first condition. Now let's test the next condition. That is, let's try to figure out if our function is differentiable, OK? Differentiable, let me write that properly. On the open interval, so from 0 to 5. Now, to know if it's differential, but first we need to differentiate FF X. Now we know that F of X equals the square root of x multiplied by 5 minus X, which we could rewrite as X multiplied by 5 minus X or raise to 1/2. Now we can tell that this is a composite function, so we can differentiate it using the chain rule. So that means it's derivative is gonna be equal to. A half multiplied by X multiplied by 5 minus X rates to the point of negative or 1/2. We differentiate the alter function multiplied by the derivative of our inner function. The derivative of X multiplied by 5 minus X. Now we can find the derivative of the inner function using the product rule since this is our product. And now since it's derivative. OK Means that we're gonna keep the first differentiate the other. So for our first term, it's gonna be X multiplied by the derivative of uh 5 minus X, OK, which is gonna be -1. Plus, next we differentiate the first keep the other, so the derivative of X is 1 multiplied by 5 minus X. I know when we expand that, that's negative X plus 5 minus X. So the derivative is 5 minus 2 X, OK. Then that means the derivative of F of X will be equal to 1/2 multiplied by x multiplied by 5 minus X to the power of negative 1/2 multiplied by 5 minus 2 X. And we can even rewrite that as 5 minus 2 X. Divided by 2 multiplied by the square root of x multiplied by 5 minus X, so this is the derivative of F of X. So here we can tell that the derivative exists and it is continuous for all X in that home interval from 0 to 5. And since the expression inside the square root is positive as we've established earlier for X inside the interval, OK, that means the derivative is well defined. The derivative is well defined. And differentiable, OK. On the open interval. So now, since our derivative is differentiable under the open interval and continuous under closed interval, that means it satisfies the criteria for the mean value theorem. Therefore, A is the correct answer. Thanks a lot for watching everyone. I hope this video helps.

Checking the Mean Value Theorem

Which of the functions in Exercises 7–12 satisfy the hypotheses of the Mean Value Theorem on the given interval, and which do not? Give reasons for your answers.

f(x) = √(x(1 − x)), [0, 1]

Key Concepts

Mean Value Theorem

Continuity

Differentiability

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