- Difficulty
- Introductory
- Suggested time
- 2 min
- Marks
- 1
- Theory
- CLP-1 §2.13Key results
The Mean Value Theorem practice question 1
The Mean Value Theorem guarantees a point in with for three of the functions below. For which one does it make no guarantee?
The theorem needs continuous on and differentiable on . Which function has a point in where it is not differentiable?
Test each function at , the centre of the interval: compare its slope just to the left and just to the right. Where the two disagree, the derivative does not exist.
Theory CLP-1 §2.13Key results
Worked solution
Strategy
Check the hypotheses: continuity on [−1, 1] and a derivative at every point of (−1, 1). One failure inside the interval voids the guarantee.
Working
x², sin x and eˣ are continuous and differentiable on the whole real line, so the MVT applies to each on [−1, 1].
The hypothesis fails, so the theorem is silent. Indeed the secant slope is (1 − 1)/2 = 0 and |x| never has slope 0.
Answer
The answer is the absolute value function.
Check
No such c can exist for |x|: its secant slope over [−1, 1] is 0, and its slope is −1 or 1 wherever it has one.
Where this goes wrong
Assuming that a function which fails the hypotheses must also fail the conclusion, or that it must still satisfy it. When a hypothesis fails, the theorem says nothing either way; here the conclusion happens to fail too, but that is not automatic.
Leans on
Differentiability and corners, from The definition of the derivative
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