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) = x⁴ᐟ⁵, [0, 1]

Accepted Answer

Welcome back, everyone. In this problem, we want to evaluate if the function F of X equals X the power of 89 on the interval 03 meets the criteria of the mean value theorem. A says the function satisfies the hypothesis of the mean value theorem, while B says it does not. Now what are these or what is this criteria that our function needs to meet for the mean value theorem. Well, recall, OK, that a function is. Sorry, well, let's write that another way. Recall that if. A function is. Continuous. And differentiable. OK. On a given interval, OK, continuous on the closed interval and differentiable on the open interval, then it satisfies, OK, the criteria of the mean value theorem. So if we want to show that it satisfies the hypotheses, we need to show that our function is continuous and differentiable. So, let's start with the continuity, OK? We want to show that our function is continuous on the closed interval uh 03, OK? Now how can we show that? Well, notice that F of X is a power function and power functions of the form F of X equals X to the power of P are continuous for all X greater than R equal to 0, where P is a positive real number, OK? So since F of X equals x to the power of P. Is continuous, OK. That's yeah that properly here. Continuous. On Oh sorry, 4, like we said earlier. R X greater than equal to 0? When P is a positive real number, so P is greater than 0, OK. And P in this case equals 89, so it's positive and a real number then. If IX It's continuous, OK. On our closed interval, continuous on. Are closed interval 03, so we know that it is continuous. Let's test the next part of our theorem. Is our function differentiable on the open interval? Well, what do we know about a function that is differentiable? Well, If it is differentiable on the open interval, then if we were to find the derivative of FF X and It satisfies the condition on that interval. That means it in conjunction with the property of continuity would satisfy the mean value theorem. So if we're gonna do anything, we need to differentiate F of X. Now we know that F of X equals X the pore of 89, so the derivative of F of X by the poor rule will be equal to 89 multiplied by x to the pore of 89 minus 1, which is -19, or we could rewrite that as 8 divided by 9 multiplied by x to the power of 19th. Now, for our function here, is it differentiable on the open interval? OK, is it differentiable on, sorry, 03? Well, if we think about our function, OK. If we, if the derivative exists. OK, the derivative would exist. If X is greater than 0 is not equal to 0, sorry. X is not equal to 0, because if X is equal to 0, then our denominator would be 0 and thus our function would be undefined. No, no, here. That for our interval it goes from 0 to 3, so it does not include 0, OK? So, because the MVT only requires differentiability on the open interval from 0 to 3, not at the end point where X equals 0, that means F of X is differentiable. FF X is differentiable on 03 because it does not include 0. Therefore, since the function is differentiable on the open interval and continuous on the closed interval, then that means the function satisfies the hypothesis of the mean value theorem. A is the correct answer. Thanks a lot for watching, everyone. I hope this video helped.

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) = x⁴ᐟ⁵, [0, 1]

Key Concepts

Mean Value Theorem

Continuity

Differentiability

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