- Difficulty
- Standard
- Suggested time
- 3 min
- Marks
- 2
- Theory
- CLP-1 §2.13Key results
The Mean Value Theorem practice question 4
Find the point promised by the Mean Value Theorem for on the interval .
Which two slopes does the theorem set equal on , and what is the derivative of ?
The secant slope is . Set equal to it.
and .
An exact expression, or a decimal to two decimal places.
Theory CLP-1 §2.13Key results
Worked solution
Strategy
Compute the secant slope using the exact values of log at the endpoints, then solve 1/c equals that slope.
Working
Secant slope.
Solve. Check: 1 < e − 1 ≈ 1.72 < e ≈ 2.72, inside the interval.
Answer
The answer is e minus 1, about 1.718.
Check
The tangent slope at 1.718 is 1/1.718 ≈ 0.582, the same as the chord from (1, 0) to (e, 1), which rises 1 over a run of e − 1.
Where this goes wrong
Writing f′(c) = log c instead of 1/c, or forgetting that log 1 = 0. The derivative of the logarithm is the reciprocal.
Leans on
The derivative of log x, from The chain rule and logarithms
Try a similar question
Find the value of in whose existence the Mean Value Theorem guarantees for on .Saved in this browser only.