(−1, 0) into the second equation, we get 0 = c(−1) + (−1)^2 , which simplifies to c = 1. Now wecan find the values for a, b, and c. We get a = 1, b = 0, and c = 1.SOLUTIONS TO PRACTICE PROBLEM SET 9
- c = 0
The Mean Value Theorem says that: If f(x) is continuous on the interval [a, b] and isdifferentiable everywhere on the interval (a, b), then there exists at least one number c on theinterval (a, b) such that f′(c) = . Here, the function is f(x) = 3x^2 + 5x − 2 and theinterval is [−1, 1]. Thus, the Mean Value Theorem says that f′(c) = . This simplifies to f′(c) = 5. Next, we need to find f′(c). The
derivative of f(x) is f′(x) = 6x + 5, so f′(c) = 6c + 5. Now, we can solve for c: 6c + 5 = 5 and c= 0. Note that 0 is in the interval (−1, 1), just as we expected.- c =
The Mean Value Theorem says that: If f(x) is continuous on the interval [a, b] and isdifferentiable everywhere on the interval (a, b), then there exists at least one number c on theinterval (a, b) such that f′(c) = . Here the function is f(x) = x^3 + 24x − 16 and theinterval is [0, 4]. Thus, the Mean Value Theorem says that f′(c) = . This simplifies to f′(c) = 40. Next, we need to find f′(c) from
the equation. The derivative of f(x) is f′(x) = 3x^2 + 24, so f′(c) = 3c^2 + 24. Now, we can solvefor c: 3c^2 + 24 = 40 and c = ± . Note that is in the interval (0, 4), but − is not in theinterval. Thus, the answer is only c = . It’s very important to check that the answers you getfor c fall in the given interval when doing Mean Value Theorem problems.