What is the value of c as per the Mean Value Theorem for f(x) = |x| over the…
What is the value of c as per the Mean Value Theorem for f(x) = |x| over the interval [-8, 6]?
Answer: C. not determined — Mean Value Theorem (MVT): If a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at…
- A.
-1/7
- B.
0
- C.
not determined
- D.
1
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Correct answer: C
Mean Value Theorem (MVT): If a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) such that f'(c) = (f(b) − f(a)) / (b − a).
Here a = −8, b = 6, and f(x) = |x|.
Compute the secant slope: (f(b) − f(a))/(b − a) = (|6| − |−8|)/(6 − (−8)) = (6 − 8)/14 = −1/7.
Check the differentiability hypothesis: f(x) = |x| is not differentiable at x = 0, since its left-hand derivative is −1 and its right-hand derivative is +1 there.
Since 0 lies inside the open interval (−8, 6), f fails to be differentiable on all of (a, b), so the Mean Value Theorem's hypothesis is not satisfied.
Even setting aside the failed hypothesis, f'(x) only ever equals −1 (for x < 0) or +1 (for x > 0); it never equals the secant slope −1/7 for any x. This directly confirms that no point c can satisfy the theorem's conclusion.
Because the differentiability condition breaks down at x = 0 inside (−8, 6), the Mean Value Theorem cannot be applied here, and the value of c cannot be determined.