Math · Calculus I · Worked example
When the Mean Value Theorem does not apply
Does the Mean Value Theorem apply to f(x) = |x| on [−1, 2]? Is there a c with f′(c) equal to the secant slope?
Find the secant slope
The endpoints are (−1, 1) and (2, 2).
Look at the derivative
f′(x) = −1 for x < 0 and 1 for x > 0, and f′(0) does not exist. No c gives 1/3.
Find the failed condition
|x| is continuous on [−1, 2] but not differentiable at 0, which is inside the interval. The theorem does not apply, and here its conclusion fails.
Result
No. |x| is not differentiable at 0, and no c has f′(c) = 1/3.
Your turn
Does Rolle’s theorem apply to f(x) = 1/x² on [−1, 1]?
Show the answer and explanation
No: f is not continuous at 0.
f(−1) = f(1) = 1, but 1/x² is undefined at 0, inside the interval. Indeed f′(x) = −2/x³ is never 0.
Keep exploring
In Graph, no tangent to y = |x| is parallel to the secant through (−1, 1) and (2, 2). On [1, 2] the theorem applies: the secant slope is 1, which every tangent there matches.
Return to the concept →Sources and scope
Authored study material. Tool results depend on the stated inputs and model assumptions.
Try in the workspace
Open the example inputs, change a value and keep a useful result on your board.
See the corner in Graph Open worked example on a board Mean Value Theorem in Math ReferenceYour existing work stays on this device. Examples open as editable copies.