Question:
The value of C of the Cauchy's mean value theorem for the function $f(x)=e^{x}$ and $g(x)=e^{-x}$ defined on $[a,b]$ is}
The value of C of the Cauchy's mean value theorem for the function $f(x)=e^{x}$ and $g(x)=e^{-x}$ defined on $[a,b]$ is}
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For exponential functions, the Mean Value is usually the arithmetic mean of the endpoints.
Updated On: Jul 3, 2026
- ab
- $\frac{a+b}{2}$
- $a+b$
- $a-b$
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The Correct Option is B
Solution and Explanation
Step 1: Concept Cauchy's Mean Value Theorem states: $\frac{f'(c)}{g'(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}$.
Step 2: Meaning Derivatives: $f'(x) = e^x$ and $g'(x) = -e^{-x}$. So, $\frac{f'(c)}{g'(c)} = \frac{e^c}{-e^{-c}} = -e^{2c}$.
Step 3: Analysis Right side: $\frac{e^b - e^a}{e^{-b} - e^{-a}} = \frac{e^b - e^a}{\frac{1}{e^b} - \frac{1}{e^a}} = \frac{e^b - e^a}{\frac{e^a - e^b}{e^a e^b}} = -(e^a e^b) = -e^{a+b}$.
Step 4: Conclusion Equating both sides: $-e^{2c} = -e^{a+b} \Rightarrow 2c = a+b \Rightarrow c = \frac{a+b}{2}$. Final Answer: (B)
Step 2: Meaning Derivatives: $f'(x) = e^x$ and $g'(x) = -e^{-x}$. So, $\frac{f'(c)}{g'(c)} = \frac{e^c}{-e^{-c}} = -e^{2c}$.
Step 3: Analysis Right side: $\frac{e^b - e^a}{e^{-b} - e^{-a}} = \frac{e^b - e^a}{\frac{1}{e^b} - \frac{1}{e^a}} = \frac{e^b - e^a}{\frac{e^a - e^b}{e^a e^b}} = -(e^a e^b) = -e^{a+b}$.
Step 4: Conclusion Equating both sides: $-e^{2c} = -e^{a+b} \Rightarrow 2c = a+b \Rightarrow c = \frac{a+b}{2}$. Final Answer: (B)
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