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) = ln 2x; [1,e]

Accepted Answer

Hi everyone. Let's take a look at this practice problem dealing with the mean value theorem. This problem says analyze the function H of X is equal to the natural log of the quantity of 4 X in quantity over the closed interval from 2 to E cubed. 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 the specified interval. And for Part B, if applicable, determine the points that the mean value theorem guarantees. We're given 4 possible choices as our answers. For choice A, for part A we have yes. For Part B, we have C is equal to the quantity of EQ plus 2 in quantity, divided by the quantity of 3 minus the natural log of 2 in quantity. For choice B, for Part A we have yes. For Part B, we have C is equal to the quantity of E cube plus 2 in quantity, divided by the quantity of 3 plus the natural log of 2 in quantity. For choice C for part A we have yes. For Part B, we have C is equal to the quantity of E cubed minus 2 in quantity, divided by the quantity of 3 minus the natural log of 2 in quantity. And for choice D for part A, we have no, and for part B, we have does not exist. Now, for part A, we need to uh verify that the mean value theorem is applicable to our function on the specified interval. To recall, to verify that the mean value theorem is applicable to a function or a specified in full, we need to satisfy two conditions. The first condition is that our function H of X is going to be continuous. On the closed interval from 2 to a cubed. And the second condition that we need to satisfy, is that H of X has to be differentiable. On the open interval from 2 to ecute. Now, if we look at our function for HF X, we see that it is equal to the natural log of the quantity of 4 X in quantity. So here, HF X is a composition of two continuous and differentiable functions, the natural log function as well as a linear function. And so that means that since it's a composition of a differentiable and Uh, continuous functions, that means that H of X is also continuous and differentiable. So H of X is going to be continuous on the closed interval from 2 to E cubed. That means our first condition is met, and it's going to be differentiable on the open interval from 2 to E cubed. So our second condition is met. So that means that our mean value theorem is applicable to our function on the specified interval. So the answer for part A is yes, the mean value theorem is applicable. Now for part B, We need to apply our mean value theorem to determine the points that that theorem guarantees. So recall that the mean value theorem states that there exists a point C in the open interval from 2 to E cubed, such that H C is equal to. H of ecute. Minus H of 2. And that quantity is divided by the quantity of E cubed minus 2. So we're gonna need to calculate several different quantities in order to be able to evaluate this expression. So first we need to determine the value of H of 2. So H of 2 is going to be equal to the natural log of the quantity of 4 multiplied by 2 in quantity, which is equal to the natural log of 8. However, we can rewrite this as being equal to the natural log of 2Q. and using our properties of logarithms, this is going to be equal to. 3 multiplied by the natural log of 2. Now, we also need to evaluate H of E cubed. This is going to be equal to the natural log of the quantity of 4 multiplied by E cubed. So, we use our properties of logarithms to separate this out into two different logarithms. So this is going to be equal to the natural log of 4. Plus the natural log of E cut. And this is going to be equal to. For the first term, we can rewrite it as the natural log of 2 squared. Then we'll have plus, for the second term, we can use our properties of logarithms to rewrite this as 3 multiplied by the natural log of E. And this is going to be equal to. For the first term, we'll use our properties of logarithms to rewrite that as 2 multiplied by the natural log of 2. Plus, for the second term, the natural log of E is equal to 1, and we'll have 3 multiplied by 1, so we'll just have 3 for the second term. And then finally, we need to calculate our derivative of H with respect to X. So we need to calculate H of X. Which is equal to the derivative with respect to X of the natural log of the quantity of 4 X. So taking the derivative, this is going to be equal to. When I take the derivative of the natural log function, I get 1 divided by the argument, so I'll have 1 divided by 4 X. But this quantity needs to be multiplied by the derivative of our argument. So basically applying our chain rule. When I take the derivative of 4 X with respect to X, I get 4. So I'll multiply 1 divided by 4 X by 4. And some find this expression, this is just going to be equal to 1 divided by X. So now we need to substitute all this information back into our expression for H C. In doing so, I'm gonna have 1 divided by C. Equal to the quantity of 2 multiplied by the natural log of 2 + 3 minus 3 multiplied by the natural log of 2 in quantity, divided by the quantity of E cubed minus 2. Now, to solve this for C, I just need to invert both sides of this equation. So I will get C is equal to E cubed minus 2. And that whole quantity is divided by the quantity, and here we simplify our expression, and then it's going to be 3 minus the natural log of 2. And if we evaluate this expression in our calculator, we'll see that it has a value that is approximately equal to 7.8. So, it does fall within the open interval from 22 E cubed. So this here is going to be our answer for Part B. Then I'll have both part answers. I can compare those to our answer choices, and we see that the correct answer choice corresponds to answer choice C. So I hope that this explanation has been useful, and I'll see you in the next video.

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