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

Step 1: Recall the Mean Value Theorem.
For \(f\) on \([e^2,e^3]\) there is a point \(c\) with \(f'(c)=\dfrac{f(e^3)-f(e^2)}{e^3-e^2}\). We just compute both sides.

Step 2: Find the derivative.
Here \(f(x)=\log x\), so \(f'(x)=\dfrac1x\) and therefore \(f'(c)=\dfrac1c\).

Step 3: Compute the average slope.
\(f(e^3)=\log e^3=3\) and \(f(e^2)=\log e^2=2\), so the numerator is \(3-2=1\).

Step 4: Write the right hand side.
\[\frac{f(e^3)-f(e^2)}{e^3-e^2}=\frac{1}{e^3-e^2}.\]
Step 5: Equate the two sides.
\[\frac1c=\frac{1}{e^3-e^2}.\]
Step 6: Solve for c.
Taking reciprocals, \(c=e^3-e^2\), which lies between \(e^2\) and \(e^3\). This is option 4.
\[ \boxed{c=e^3-e^2} \]