Answer: First, find f(0) and f(1).
f(0) = 0^2 + 2(0) − 1 = −1 and f(1) = 1^2 + 2(1) − 1 = 2Then,
= = 3 = f′(c)Next, find f ́(x).
f ́(x) = 2x + 2Thus, f ́(c) = 2c + 2 = 3, and c = .
PROBLEM 2. Find the values of c that satisfy the MVTD for f(x) = x^3 + 1 on the interval [1, 2].
Answer: Find f(1) = 1^3 + 1 = 2 and f(2) = 2^3 + 1 = 9. Then,
= 7 = f′ (c)Next, f′(x) = 3x^2 , so f′(c) = 3c^2 = 7 and c = ±.
Notice that there are two answers for c, but only one of them is in the interval. The answer is c =
PROBLEM 3. Find the values of c that satisfy the MVTD for f(x) = x + on the interval [−4, 4].
Answer: First, because the function is not continuous on the interval, there may not be a solution for c.
Let’s show that this is true. Find f(−4) = −4 − = − and f(4) = 4 + = . Then,
Next, f′(x) = 1− . Therefore, f′(c)= 1− = .
There’s no solution to this equation.
PROBLEM 4. Find the values of c that satisfy Rolle’s Theorem for f(x) = x^4 − x on the interval [0, 1].