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) = x²ᐟ³, [−1, 8]

Accepted Answer

Welcome back, everyone. Evaluate if the function H of X equals X the power of 4/7 on the interval from -2 to 11 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 do we know about this criteria for the mean value theorem? Well, in order for it to satisfy the mean value theorem, OK, recall that the function H of X. Or any function in general. Has to be continuous, OK. And the friendship. OK. On the given interval. To satisfy the mean value theorem, yes, to satisfy MVT. So now, in our case, that means we need to show that the function is continuous and that it's differentiable on the entire interval. So let's try to do that. First, let's try to prove that it is continuous, OK, so let's tackle its continuity. Now if we think about it here, if we look at H of X, it's equal to x to the power of 4/7. It's a power function with a rational exponent. Now power functions of the form X ton are continuous for all real numbers, even when N is a rational number, provided that X is greater than or equal to 0 or that it's negative, as long as the exponent is an integer or a rational number with an odd denominator. Now for our function HFX, OK. Notice that this is a positive. A rational number. It's a positive rational number. So since it satisfies the criteria, OK, then it is continuous. It is continuous on all of the interval, OK? Continuous on all of the interval from -2 to 11 including negative values of X, so it satisfies the first condition. Now the question is, does it satisfy the second condition? Is it differentiable on the entire interval? Well, first, let's find out. So let's call that differentiability. Yeah. In order to figure that out, let's first differentiate F of X. Can it be differentiated? Well, sorry, HFX, my apologies. No. Since H of X is X of 4/7, then by the poor rule, its derivative is going to be equal to 4/7 of X of 14/7 minus 1, which is negative 3/7. So we could rewrite the derivative of H of X as 4/7 multiplied by 1 divided by X to the power of 3/7. No, for the derivative to be defined, X must not be 0, OK, because our denominator would be undefined at X equals 0, OK. No, the question is, is it possible for X to be zero in this interval? Well, Uh let's, let's, let's write that out here. So since X equals 0 is in the interval, because the interval goes from -2 to 11, so it includes 0, OK. Then our denominator, X 37. Would be undefined. OK, because you know, uh 0 to 37 would be 0, and if our denominator is 0, the function is undefined. OK. So since it's undefined, it implies that the function H of X is not differentiable. At X equals 0. It's differentiable at all the other points, but it's not differentiable at X equals 0. So in other words, our function satisfies one condition, it's continuous, but it doesn't satisfy the other. It is not differentiable at least on one point on the interval. So since it's continuous but not, it's continuous on the close interval but not differentiable on the old interval, that means the function does not satisfy the hypothesis of the mean value theorem on our interval. Therefore, B. 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) = x²ᐟ³, [−1, 8]

Key Concepts

Mean Value Theorem

Continuity

Differentiability

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