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) = x + 1/x; [1,3]

Accepted Answer

Hello. In this video, we are going to be answering the following question using the mean value theorem. We are told to analyze the function H of X equal to X cubed minus 1 divided by X on the closed interval from 1 to 4. The first part of the problem asks us to decide whether the mean value theorem can be applied to this function on the specific interval. So, in order to approach this problem, let's just go ahead and list out the conditions for the mean value theorem. The mean value theorem, MTV for short, has three specific conditions. First, is the graph continuous on the closed interval from 1 to 4? Second, Is it differentiable? On the open interval from 1 to 4, and if the first two conditions are true, then there is a third condition that states that H of C there will be an average value c within the interval that is equal. To H of 4 minus H of 1 divided by 4 minus 1. So, in order to answer part A of this problem, whether to decide that mean value theorem can be applied to this problem, we first need to check to see if the first two conditions of the mean value theorem are true. So, let's go ahead and start with condition one. Is H of X continuous? On the closed interval from 1 to 4. Now, in order to answer this question, we can go ahead and check the domain of the given function HX. Now, again, H of X is given to us as X cubed minus 1 divided by X. Now, H of X is a has a difference of a polynomial and a rational function. Now we know that the domain of the polynomial is all real numbers, from negative infinity to positive infinity. But the problem lies in the second term with the rational function. The value of X that causes the rational function to be undefined is when X is equal to 0. We cannot have division by 0, otherwise we will have an undefined value. What this means is that the restriction on the domain is that X cannot equal to 0. Now, if we look at our interval 1 through 4, is 0 located between the values of 1 through 4? The answer to that problem is no, 0 is not contained in 1 through 4, and that means that this graph is going to be continuous on the interval from 1 to 4, since the only restriction is not contained in the interval. So the answer to the first condition is yes. Now, what about the second condition? The second condition asks us to check, is HX differentiable? On the open interval from 1 to 4. In order to check differentiability, what we first need to do is we need to calculate the derivative of H of X and then decide whether the restrictions of the derivative are continuous on the open interval from 1 through 4. So, let's calculate the derivative H X. Now, using the power rule, the derivative of X cubed will be 3 X squared. Then we will subtract the derivative of 1 divided by X. The derivative of 1 divided by X will be negative 1 divided by X2d through the quotient rule. Multiplying the two negative terms together will give us the derivative 3 X2 plus 1 divided by X. Now, if we compare this derivative to the original function, notice that we will end up getting the same restriction X cannot equal to 0. Since 0 is not contained in the open interval 1 through 4, that means that H of X is differentiable at every value between 1 through 4. So, the second condition is true. Now that we've checked the first two conditions of the mean value theorem, we can now assume that there is a value C. Such that H C is equal to the average rate of change between the interval, that will be H of 4 minus H 1 divided by 4 minus 1, and that is going to lead into part B of this problem. Part B asks us, if we can apply the mean value theorem, we want to identify the points that are guaranteed to exist by the mean value theorem. So what this means is that we are going to use the 3rd condition to find the values of C. Now, before we start finding the values of C, we first need to calculate the average rate of change through the mean value theorem. Now recall that the given function is H X equal to X cubed minus 1 divided by X. In order to obtain H of 4, we will be plugging 4 into the original function. That will give us 4 cubed. -1 divided by 4. Then we will subtract H of 1, which will give us the quantity 1 cubed minus 1 divided by 1, and this will all be divided by 4 minus 1, which is 3. Now the numerator will simplify. To be 63.75. Minus 0 divided by 3, and this will give us the value of 21.25. So what this tells us is that there is a value C H C that is going to give us the value 21.25. But this doesn't tell us what the value C is going to be. Instead, what we need to do is we now need to resolve the left hand side. We are going to take the variable C and plug it into the derivative that we solved for earlier. Now, recall that that derivative was 3 X2 plus 1 divided by X. That will allow us to rewrite the left-hand side of this equation to be 3 C 2 plus 1 divided by C2 equal to 21.25. After algebraically simplifying and solving for C, that will give us a value of C. Approximately equal to 2.7, which is going to be the solution to Part B of the problem. So just to recap, we answered part A by checking to see if H of X is continuous on the closed interval from 1 to 4, and if it is differential on the open interval 1 through 4. We found that both of those are true. And by using the second condition for Part B, we solved that the guaranteed value from the mean value theorem is going to be 2.7. With that being said, the solution to this problem is going to be B. So, I hope this video helps you in understanding on how to use the mean value theorem, and I will go ahead and see you all in the next video.

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