Checking the Mean Value Theorem

Question

Find the value or values of c that satisfy the equation (f(b) − f(a)) / (b − a) = f′(c) in the conclusion of the Mean Value Theorem for the functions and intervals in Exercises 1–6.

g(x) = {x³, −2 ≤ x ≤ 0

x², 0 < x ≤ 2

Accepted Answer

Calculate the value of C that satisfies the equation K of B minus K A divided by B minus A equals K C for the function K of X, which is defined as K X equals 2 XQ plus 12 from -1 to 2, and 6X2 + 4 from 2 to 3. We have 4 possible answers, being 0, square root of 22, and 1. Now, we first noticed that this is the form of a mean value theorem. So, we need to check continuity and differentiability before we can solve for C. So for continuity. We're going to check the left limit and the right limit at X equals 2. Because that's the point they have in common. So limit as X approaches to from the left will be 2 multiplied by 2 cubed plus 12. Which is 28. Also need the check that you write, limit. So we have the limit, as X approaches to from the right. Which is 6, multiplied by 2 squad, plus 4. This is 28. Our continuity check passes. Now we need to check differentability. So we need to check if both functions are differentiable over the open interval. We have 2 X cubes plus 12. Which is differentiable because it's a polynomial. And the same for 6 X2 + 4. So we can confirm that this satisfies the mean value theorem. Now we can move on to use our equation. We have OK. Of B minus K of A divided by B minus A equals K. Of C Now B is our largest value of our interval. And if you look, that value is 3, so BFKF 3. Minus K of -1 divided by 3 minus -1. Now we can solve KF 3 is going to go with the 2nd function. We have 6 multiplied by 3 squared. Plus 4. Which is 58. K of negative 1 goes with the first function. Which is 2 multiplied by -1. Cubed plus 12 or just. Now, we can solve. We get 58 minus 10, divided by 4. Which is 48 divided by 4, or 12. Now we need to find K of C. K C We can look at our function. And notice That we're looking for for K C equals 12. This then gives us 6 C 2 equals 12, based on our function. This is for the interval. From 1 to 2. Or C is in Now we just solve. Divide both sides by 6 to get C2 equals 2. And C equals plus or minus the square root of 2. So then we notice. That We'll take the positive value. Because square root of 2 is not in the interval of -12. We'll also check our other function. Which is From 2 to 3. We get 12 seats. Equals 12 Or C equals 1. This is from the interval. From 2 to 3. C is not. In the interval of 2 to 3, because C equals 1. That means that we won't use the value for C equals 1. This means that we only have one possible answer, which is C equals the square root of 2. We can see the answer to our problem is answer B. OK, hope to help you solve the problem. Thank you for watching. Goodbye.

Checking the Mean Value Theorem

Find the value or values of c that satisfy the equation (f(b) − f(a)) / (b − a) = f′(c) in the conclusion of the Mean Value Theorem for the functions and intervals in Exercises 1–6.

g(x) = {x³, −2 ≤ x ≤ 0
x², 0 < x ≤ 2

Key Concepts

Mean Value Theorem

Piecewise Functions

Differentiability and Continuity

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