- c =
The Mean Value Theorem says that: If f(x) is continuous on the interval [a, b] and is
differentiable everywhere on the interval (a, b), then there exists at least one number c on the
interval (a, b) such that f′(c) = . Here the function is f(x) = − 3 and the interval
is [1, 2]. Thus, the Mean Value Theorem says that f′(c) = . This simplifies to f′
(c) = −3. Next, we need to find f′(c) from the equation. The derivative of f(x) is f′(x) = − , so
f′(c) = − . Now, we can solve for c: − = −3 and c = ± . Note that c = is in the
interval (1, 2), but − is not in the interval. Thus, the answer is only c = . It’s very
important to check that the answers you get for c fall in the given interval when doing Mean
Value Theorem problems.
- No Solution.
The Mean Value Theorem says that: If f(x) is continuous on the interval [a, b] and is
differentiable everywhere on the interval (a, b), then there exists at least one number c on the
interval (a, b) such that f′(c) = . Here the function is f(x) = − 3 and the interval
is [−1, 2]. Note that the function is not continuous on the interval. It has an essential
discontinuity (vertical asymptote) at x = 0. Thus, the Mean Value Theorem does not apply on
the interval, and there is no solution.
Suppose that we were to apply the theorem anyway. We would get f′(c) = .
This simplifies to f′(c) = 3. Next, we need to find f′(c) from the equation. The derivative of f(x)
is f′(x) = − , so f′ = − . Now, we can solve for c:− = 3. This has no real solution.
Therefore, remember that it’s very important to check that the function is continuous and
differentiable everywhere on the given interval (it does not have to be differentiable at the
