Verla
MathCalculus

Calculus Theorems: Why the Mean Value Theorem Fails at a Corner

Prove that the derivative never matches the secant slope for a shifted absolute-value function, and identify the failed Mean Value Theorem hypothesis.

Question

For

f(x)=∣x−12∣,f(x)=\left|x-\frac12\right|,

show that no cc in (−1,1)(-1,1) satisfies

f(1)−f(−1)=2f′(c).f(1)-f(-1)=2f'(c).

Explain why the Mean Value Theorem does not apply on [−1,1][-1,1].

This question has been answered.

Explore the full answer and supporting references to better understand this question.

Verla BasicBest value

$19.99/month

  • Everything in Study Pass
  • Assignments with writing output
  • AI detection and humanizer
  • Basic quotas across every tool
Study Pass

$0.99/month

  • Entire question library
  • Knowledge guides and sources
  • Detailed answers and explanations
  • No AI tools or quotas

Evidence boundary

This solution addresses OpenStax Calculus Volume 1 section 4.4, exercise 167, with f(x) = |x − 1/2| on [−1,1]. The no-c conclusion follows from the exact derivative values and the undefined derivative at the corner. It is not inferred merely from failure of a theorem hypothesis.