Means

b. Show that the point guaranteed to ex...

Question

Means

b. Show that the point guaranteed to exist by the Mean Value Theorem for f(x) = 1/x on [a,b], where 0 < a < b, is the geometric mean of a and b; that is, c = √ab.

Accepted Answer

Hello. In this video, we are going to be approaching the following mean value theorem problem. We are asked to consider the function F of X equal to X2 on the interval from 2 to 5. According to the mean value theorem, there exists at least 1 point C on the open interval from 2 to 5, such that F equals the average rate of change of the function from the interval from 2 to 5. What is the value of C? Well, before we approach this problem, let's just go ahead and write down what the problem tells us. Now, first, we are told that we have a function F of X equal to X2, and we are told to analyze this function on the closed interval from 2 to 5. Now, by the mean value theorem, since it guarantees a point C that gives us the average rate of change of X, we want to find this value of C. In order to use the mean value theorem, we first need to check the first two conditions of the mean value theorem. The first condition asks us, is F of X continuous? On the interval, the closed interval from 2 to 5. In order to answer this question, we can check the domain of the given function over the interval. F of X equal to X2 is a function of a polynomial. The one thing we know about polynomial function is that its domain is all real numbers from negative infinity to positive infinity. What this means is that the function will be continuous on the closed interval from 2 to 5. So, the first condition of the mean value theorem is satisfied. Now we must check at the 2nd condition. The second condition asks us, is F of X differentiable? On the open interval from 2 to 5. In order to check this condition, we can take the derivative of F of X and see if the domain of F of X is continuous on the open interval from 2 to 5. So, by the power rule, F of X will equal to 2 X. Again, this is in the form of a polynomial function. Its domain will be all real numbers, which means that its domain will be continuous on the open interval from 2 to 5. This verifies that F of X is differentiable on the open interval from 2 to 5. Now that the first two conditions are satisfied, the third condition tells us. That there is a value F of C, that gives us the average rate of change of F of X over the interval from 2 to 5. That is equal to F of 5 minus F of 2, divided by 5 minus 2. So, our goal here is to solve for this value of C. In order to start this problem, let's first simplify the right-hand side of this equation. So, we will go ahead and plug in F of 5, which is going to give us the quantity of 5 squared equal to 25. Next, we need F of 2, and plugging this into F of X will give us 2 squared equal to 4. That will allow us to simplify this equation to be 25 minus 4, divided by 5 minus 2, which is 3. The numerator will further simplify to be 21 divided by 3, which will finally simplify to be 7. What this tells us is that F of C is equal to 7. Now, what we need to do is we need to plug in the variable C into the derivative of F of X in order to solve for the value of C that gives us 7. Again, we solve for this derivative. The derivative F of X is equal to 2 X. So if we plug inc to the derivative, we can rewrite this equation to be 2 C equal to 7, and by dividing both sides of this equation by 2, we get C is equal to 7 divided by 2. That tells us that the mean value theorem guarantees that 7 divided by 2 will give us the average rate of change of 7, and this is going to be the solution to the problem. With that being said, the overall solution is going to be D. So, I hope this video helps you in understanding how to use the mean value theorem, and I will go ahead and see you all in the next video.

Meansb. Show that the point guaranteed to exist by the Mean Value Theorem for f(x) = 1/x on [a,b], where 0 < a < b, is the geometric mean of a and b; that is, c = √ab.

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Means

b. Show that the point guaranteed to exist by the Mean Value Theorem for f(x) = 1/x on [a,b], where 0 < a < b, is the geometric mean of a and b; that is, c = √ab.