Question

For the function \(f(x)=\log x\), the number \(c\) strictly between \(e^2\) and \(e^3\) that satisfies \(f'(c)=\dfrac{f(e^3)-f(e^2)}{e^3-e^2}\) is

• A. \(\dfrac{1}{e^3 - e^2}\)
• B. \(e^2 - e^3\)
• C. \(\dfrac{1}{e^2 - e^3}\)
• D. \(e^3 - e^2\)

Hint:

This is the Mean Value Theorem. It says there is a point \(c\) where the slope of the tangent equals the slope of the line joining the two end points.

Answer 1

The Correct Option is D

Concept:
This is the Mean Value Theorem. It says there is a point \(c\) where the slope of the tangent equals the slope of the line joining the two end points. We just set the derivative equal to that average slope and solve for \(c\).

Step 1:
The function is \(f(x)=\log x\), so its derivative is \(f'(x)=\dfrac{1}{x}\), and at \(c\) we get \(f'(c)=\dfrac{1}{c}\).

Step 2:
Compute the average slope on the right hand side. Since \(f(e^3)=\log e^3 = 3\) and \(f(e^2)=\log e^2 = 2\): \[\frac{f(e^3)-f(e^2)}{e^3-e^2} = \frac{3-2}{e^3-e^2} = \frac{1}{e^3-e^2}.\]

Step 3:
Set the two equal: \[\frac{1}{c} = \frac{1}{e^3 - e^2}.\] Taking reciprocals gives \(c = e^3 - e^2\).

Step 4:
Check it lies between \(e^2\) and \(e^3\). Since \(e^3 - e^2 = e^2(e-1)\approx 7.39\times1.72\approx 12.7\), and \(e^2\approx 7.39\), \(e^3\approx 20.1\), yes it is in between.

Answer: Option (4) — \(e^3 - e^2\).