Let ƒ(x) = x²⸍³ . Show that there is no value of c in the...

Question

Let ƒ(x) = x²⸍³ . Show that there is no value of c in the interval (-1, 8) for which ƒ' (c) = (ƒ(8) - ƒ (-1)) / (8 - (-1)) and explain why this does not violate the Mean Value Theorem.

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says consider H of X is equal to the cube root of X. Determine the values of C in the open interval from -1 to 27, at which the equation H C is equal to the quantity of H 27, minus H 1 in quantity, divided by the quantity of 27 minus 1 in quantity holds true. Provide a justification as to why the mean value theorem cannot be applied in this case. Now, the first part of this question wants us to determine the values of C in the open interval from -1 to 27. at which the equation of H C It's equal to the quantity of H of 27. Minus each of -1 in quantity, divided by the quantity of 27, minus 1 in quantity holds true. So, we need to calculate several um quantities in order to evaluate this expression. The first quantity that we need to calculate is gonna be H of -1. And we'll use our function H of X that we were given in the problem to evaluate this expression. So substituting N minus 1 for X in that equation, we'll have H of -1, is equal to the cube root of -1. Which is equal to -1. We also need to evaluate H of 27, and this is equal to the cube root. Of 27, which is equal to 3. We also need to take a derivative of our function H of X with respect to X. So, we need to calculate H of X, and this is going to be equal to the derivative with respect to X. Of the quantity of X-rays to the 1/3 power and here we've rewritten the cube root of X as X-rays to the 1/3 power, and this is to make it easier to use our power rule to take this derivative. So taking the derivative, this is going to be equal to 1 divided by 3, multiplied by x raised to the minus 2/3 power. And we can rewrite this as being equal to 1 divided by the quantity of 3, multiplied by the cube root of X2. So we'll take all this information and plug it back into our expression for HC. So doing so, I'm going to have 1 divided by the quantity of 3, multiplied by the U root of D2d. Here we replace X with C, and this is going to be equal to the quantity of H of 27, which is 3, minus H of -1, which is -1, in quantity, divided by 27 minus -1, which is 28. So simplifying the expression on the right hand side, we'll get 1 divided by the quantity of 3 multiplied by the cube root of C2 in quantity is equal to. Here I'll have 3 minus minus 1, which will give me 4, and 4 divided by 28 gives me 1 divided by 7. So now I need to solve this expression for C. So, in inverting both sides of this equation, we'll get 3 multiplied by the cube root of C2 is equal to 7. Now, to isolate C, the first step I need to do is divide both sides by 3, and so I'll have the cube root of C2d. It's equal to 7 divided by 3. Then we'll need to cube both sides of this equation. In doing so, we'll get C2d. is equal to the quantity of 7 divided by 3 in quantity cubed, then we just need to take the square root of both sides. Then we'll have C is equal to the quantity of 7 divided by 3 in quantity raised to the 3 halves power. And we can actually rewrite this as being equal to 7, multiplied by the square root of 7. Divided by the quantity of 3 multiplied by the square root of 3. So this is actually the answer for C that we were looking for. So that is the first part of our question answered. Now if we look at the second part, we need to provide a justification as to why the mean value theorem cannot be applied in this case. And for this, we need to look at our function H X. So recall for the mean value theorem to be applicable, we need our function H of X to be differentiable everywhere in the open interval in this case from -1 to 27. But if we look at our function H of X, we notice that H 0 does not exist. So I will have 1 divided by 0, which is not defined. So, here we have H of 0. Does not exist, and since it does not exist, that means our function H of X is not differentiable everywhere, and so therefore the mean value theorem does not apply. So, I hope that this explanation has been useful, and I'll see you in the next video.

Let ƒ(x) = x²⸍³ . Show that there is no value of c in the interval (-1, 8) for which ​​ƒ' (c) = (ƒ(8) - ƒ (-1)) / (8 - (-1)) and explain why this does not violate the Mean Value Theorem.

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Let ƒ(x) = x²⸍³ . Show that there is no value of c in the interval (-1, 8) for which ƒ' (c) = (ƒ(8) - ƒ (-1)) / (8 - (-1)) and explain why this does not violate the Mean Value Theorem.