Mean Value Theorem for quadratic functions Consider the quadra...

Question

Mean Value Theorem for quadratic functions Consider the quadratic function f(x) = Ax² + Bx + C, where A, B, and C are real numbers with A ≠ 0. Show that when the Mean Value Theorem is applied to f on the interval [a,b], the number guaranteed by the theorem is the midpoint of the interval.

Accepted Answer

Hello. In this video, we are going to be working on the mean value theorem problem. We are asked to apply the mean value theorem to the following function H of X on the integral little M to little N, and show that the number P guaranteed by the theorem is the midpoint of the interval. We are told that H of X is equal to MX2 + N X plus P, where M, N, and P are real numbers, and capital M is not zero. So how can we start this problem? Well, in order to apply the mean value theorem, the first thing we need to do is we need to check the first two conditions of the mean value theorem. The first condition asks us, is HX continuous? On the interval, the closed interval, little M to little N. In order to answer this problem, we can take a look at the domain of the given HFX function. Now, if we take a look at the given HFX function, notice that the function is in the form of a polynomial. The one thing we know about polynomial domains is that the domain of a polynomial is all real numbers from negative infinity to positive infinity. With this being said, H of X will be continuous on the closed interval from little M to little N. So we know that this function satisfies the first condition of the mean value theorem. Now we need to check the second condition, which asks us, is H of X differentiable? On the open interval, little M to little N. In order to check this condition, we will first take the derivative of the given HMX function and check the domain of the derivative. By the power rule, the derivative H X will equal to 2 MX plus N. Now, the derivative is in the form of a polynomial as well. And since the domain of a polynomial is all real numbers, then H of X will be differentiable on the open interval from M to N. Now that the first two conditions are satisfied, we can move on to the 3rd condition. The third condition tells us that if the first two conditions are satisfied, then there is a value P such that H P is equal to H of little N minus H of little M. Divided by little N minus little M. Now what we want to do is we want to solve for this value of P and verify that it is the midpoint of the interval. So, in order to do this, we first need to calculate our HN and our HFM function. In order to calculate H of N, we are going to plug in the variable little N into the function H of X. That will give us M multiplied by little N squared, plus N multiplied by little N plus P. We will repeat this process for H of little M. That will give us little M or M multiplied by little M squared, plus N multiplied by little M plus P. We will go ahead and take these two functions and replace them with H of N and H of M in the third condition of the mean value theorem. That will give us M multiplied by N2 plus N multiplied by little N. Plus P minus the function M multiplied by little M2 plus N multiplied by little M plus P, and this will all be divided. By N minus M. Now we need to simplify. If we distribute -1 into the function HM, And we combine like terms, we can simplify the derivative to be M and squared. Plus N multiplied by little n. Minus M multiplied by little M squared, minus M multiplied by little M, all divided by N minus M. Now let's go ahead and group up the M and N terms together. If we group up those terms together and factor out a capital M and N, we can rewrite the third condition to be capital M, multiplied by the quantity, little N squared minus M2 plus N multiplied by the quantity, little N minus little M. And this will be divided by little N minus little M. By using the difference of squares, we can rewrite the first term of the numerator to be M multiplied by the quantity, little M plus little M multiplied by little N minus little M plus N multiplied by little N minus little M, all divided. By little n minus little M. Now notice that each factor in the numerator, or each term in the numerator has a factor of N minus M. and we have that one factor in the denominator. If we factor out this quantity from the numerator and eliminate the terms from the numerator and denominator, we can simplify the statement to be capital M. Multiplied by little M plus little M plus N. This is going to be equal to H of P. Now we're not quite done yet. What we've done is we simplified the condition that was given to us by the mean value theorem, but now we need to simplify the left hand side, the right hand side in order to solve for the value of P. This is telling us to take the value of P and plug it into the derivative H. Now, earlier, in the second condition, we did calculate this derivative. We know that H of X is equal to 2 MX plus N. So, we will plug in the value P into this derivative and replace the variable X with P. That will give us the statement. 2 multiplied by capital M multiplied by P plus N is equal to M multiplied by the quantity, little N plus little M plus N. We will start by subtracting MN to both sides of the equation, leaving us with 2 MP equal to M multiplied by the quantity, little N plus little M. If we divide both sides of the equation by 2 M, that will leave us with P equal to N plus little M divided by 2. This matches the equation to a midpoint. So this verifies that the value of P is going to be the midpoint on the interval from little N to little M. And with that being said, the solution to this problem is going to be D. So, I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

Mean Value Theorem for quadratic functions Consider the quadratic function f(x) = Ax² + Bx + C, where A, B, and C are real numbers with A ≠ 0. Show that when the Mean Value Theorem is applied to f on the interval [a,b], the number guaranteed by the theorem is the midpoint of the interval.

Key Concepts

Mean Value Theorem (MVT)

Quadratic Functions

Derivative and its Interpretation

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