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) = 7 -x² ; [-1; 2]

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says to analyze the function H of X is equal to 2 X2 minus 4 X + 1 on the closed interval from 0 to 3. And there are two parts to this question. For Part A, we need to verify if the mean value theorem is applicable to this. Function on this specified interval. And for Part B, if so, identify the points that the mean value theorem guarantees. We have 4 possible choices as our answers for choice A, for Part A, we have yes. For Part B, we have C is equal to 3 divided by 2. For choice B, for Part A, we have does not apply, and for Part B, not applicable. 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 yes, and for Part B, we have C is equal to 2. Now, the first part of this question asks us to verify that the mean value theorem is applicable. And so for this to, um, this theorem to be applicable, our function needs to satisfy two conditions. One, it has to be continuous on the closed interval from 0 to 3, and it needs to be differentiable on the open interval from 0 to 3. And if we look at our function, we see that we actually have a polynomial of 2 X2 minus 4 X + 1. And polynomials are continuous and differentable everywhere on the real line. So, in this case, um, our function does satisfy both of those conditions. So that means that for part A, yes. Our mean value theorem is applicable. So, then, for Part B, we need to find the points that the mean value theorem guarantees. And so recall that the mean value states that there exists a point C, such that H of C is equal to. H of 3 minus H of 0, which is our change in H. Divided by 3 minus 0. Which is our change in our X values, has to be true. So, we need to calculate a few things. First, we need to calculate H of 0 and H of 3. So doing so, we'll have 8 of 0. is equal to? 2 multiplied by 0 squared, minus 4 multiplied by 0, plus 1, and evaluate this expression, this is going to be equal to 1. For age of 3. This is going to be equal to 2, multiplied by 3 squared. -4, multiplied by 3. Plus 1, and when we evaluate this expression, this is going to be equal to 7. Now, we also need to take a derivative with respect to H of our function H of X. So that will be H of X. This is going to be equal to the derivative with respect to X. Of the quantity of 2 X2 minus 4 X. Plus one. So taking this derivative, this is going to be equal to. 4 X minus 4. So now we can use this information in our expression. And so we'll have H of C that is going to be 4 multiplied by C minus 4. So here we just replace X with C, and this is going to be equal to H of 3 was 7, minus H 0, which is 1, and that quantity is divided by 3 minus 0, which is 3. So simplifying the expression on the right hand side, we will have 4C minus 4 is equal to 6 divided by 3. Now we need to solve this for C, so to do so, we'll need to add 4 to both sides of our equation. So we'll have 4C is equal to 6 divided by 3 is 2, and 2 + 4 is 6. And now to so for C, we'll just divide both sides by 4. And we'll get C is equal to 3 divided by 2. And so this here is our answer for Part B. So taking into account our answer of yes for part A. And looking at our answer choices, we see that the correct answer choice corresponds to answer choice A. So I hope that this explanation has been useful, and I'll see you in the next video.

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