21–32. Mean Value Theorem Consider the following functions on ...

Question

21–32. Mean Value Theorem Consider the following functions on the given interval [a, b].

a. Determine whether the Mean Value Theorem applies to the following functions on the given interval [a, b].

b. If so, find the point(s) that are guaranteed to exist by the Mean Value Theorem.

ƒ(x) = { - 2x if x < 0 ; x if x ≥ 0 ; [-1, 1]

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to analyze the function H X on the closed interval from -2 to 2. Every we given the function H of X is equal to X2d if X is less than 0 and 2 X if X is greater than or equal to 0. We get there are two parts to this question that we need to answer. For part A, we need to determine whether the mean value theorem is applicable to the function H of X on the closed interval from -2 to 2, and for part B, if applicable, identify the points guaranteed by the mean value theorem. We have 4 possible choices as our answers. Your choice A, for Part A, we have yes. For Part B, we have C is equal to 1. For choice B, for Part A, we have yes. For Part B, we have C is equal to 3 divided by 2. For choice C for Part A, we have yes. For Part B, we have C is equal to -1, and for choice D, for Part A, we have does not apply, and for Part B, we have not applicable. Now for Part A, we need to determine whether the mean value theorem is applicable. So we call that for the mean value theorem to be applicable, that our function H of X needs to be continuous on our interval. In this case, it's going to be the closed interval from -2 to 2. And our function also needs to be differentiable on the open interval from -2 to 2. So we'll start off by checking for continuity first, and we're going to focus on the point X equal to 0. Now, we call for a function to be continuous at a point that the limit as X approaches that point from both the left and the right side, must be the same. So we'll start off by looking at the limit. As X approaches 0 from the left of H of X. And this is going to be equal to the limit. As X approaches 0 from the left of X2, does X is going to be slightly less than 0. When we're approaching from the left, that means we will want to use X2d for H of X, and we can evaluate this limit by direct substitutions. This is going to be equal to 02, which is equal to 0. We'll now look at the limit, as X approaches 0 from the right of H of X. And this is going to be equal to the limit. As X approaches 0 from the right. Of 2 X, since X is gonna be slightly greater than 0, we'll want to use 2 X for H of X. And again, we can evaluate this limit by direct substitution, so this is going to be equal to 2 multiplied by 0, which is equal to 0. Now, since both of these limits of H of X evaluate to 0, That means that H of X. Is going to be continuous. So that means that the first part of our criteria is met. Now, for the second part, we need to determine whether H of X is differentiable over the entire interval. And here again, we're going to focus on X equal to 0. So we're gonna check for diff differentiability. So we're gonna need to evaluate the limit. As K approaches 0 from the left. Of the quantity and here we're using our limit definition of a derivative, and so we'll have H of 0 plus K. Minus H of 0. And that whole quantity is divided by K. So this is going to be equal to the limit. As K approaches 0 from the left. Of H of 0 plus K, it's just gonna be H of K, but here, K is going to be slightly less than 0, so that means we're gonna need to use our function for H of X as X2, so we'll have K2d. Minus H of 0, which is 0, we evaluated that previously, that is divided by K. And so simplifying this expression, this is going to be equal to the limit, as K approaches 0 from the left of K, and applying direct substitution to evaluate this limit, this is going to be equal to 0. So, we'll have to do the same um limit evaluation, except we're gonna have K um approaching 0 from the right. So this will be our right-handed derivative. So we'll have the limit as K approaches 0 from the right of H of 0 plus K. Minus H of 0, and that whole quantity is divided by K. So this is going to be equal to the limit. As K approaches 0 from the right. Of and here since K is gonna be slightly greater than 0, we're gonna have um for H of X 2 X, so we'll have 2K for H of K minus 0 for H of 0. Divided by, and that whole quantity is divided by K. So, so find this expression, this is going to be equal to the limit. As K approaches 0 from the right of 2, which is just equal to 2. I notice here that Our two limits do not evaluate to the same value. So, when we take the limit as K approaches 0 from the left, we have our derivative being equal to 0, and when K is approaching 0 from the right, our derivative evaluates to 2. So that means that H 0 does not exist. And since our derivatives does not exist at that point, that means that our function is not differentiable over the entire interval. So that means that the mean value theorem does not apply. So, since one of the conditions was not met, we cannot use the mean value theorem. So that is going to be our answer for Part A. Now, since we cannot apply our mean value theorem to this function, that means we cannot actually calculate any points for Part B. So, for Part B, it's not really applicable. So, now that we have our answer, we can compare that to our answer choices, and we see that the correct answer choice corresponds to answer choice D. So I hope that this explanation has been useful, and I'll see you in the next video.

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