For the functions f(x) = ex and g(x) = e-x on the interval [a, b], the value…
For the functions f(x) = ex and g(x) = e-x on the interval [a, b], the value of c given by Cauchy's Mean Value Theorem is:
Answer: A. (a+b)/2 — Concept: Cauchy’s Mean Value Theorem states that if f and g are continuous on [a, b], differentiable on (a, b), and g’(x) is never zero on (a, b), then there…
- A.
(a+b)/2
- B.
(a-b)/2
- C.
2a/(a+b)
- D.
2(a-b)/ab
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Correct answer: A
Concept: Cauchy’s Mean Value Theorem states that if f and g are continuous on [a, b], differentiable on (a, b), and g’(x) is never zero on (a, b), then there exists some c in (a, b) such that f’(c)/g’(c) = [f(b) − f(a)] / [g(b) − g(a)].
Application: Here f(x) = ex and g(x) = e-x, both continuous and differentiable everywhere, and g’(x) = -e-x is never zero. Applying the theorem step by step:
Differentiate: f’(x) = ex, g’(x) = -e-x, so f’(c)/g’(c) = ec / (-e-c) = -e(2c).
Compute the right-hand side: [f(b) − f(a)] / [g(b) − g(a)] = (eb − ea) / (e-b − e-a).
Simplify the denominator: e-b − e-a = (ea − eb) / e(a+b).
So the right-hand side = (eb − ea) × e(a+b) / (ea − eb) = -e(a+b).
Equate the left-hand side and right-hand side: -e(2c) = -e(a+b), so e(2c) = e(a+b), so 2c = a + b, so c = (a+b)/2.
Cross-check: substituting c = (a+b)/2 back into -e(2c) gives -e(a+b) identically, matching the right-hand side computed independently — confirming the result.
Therefore, c = (a+b)/2.