- Difficulty
- Challenging
- Suggested time
- 7 min
- Marks
- 4
- Theory
- CLP-1 §2.13Key results
The Mean Value Theorem practice question 8
Use the Mean Value Theorem to show that for every .
The right-hand side is the tangent line to at . The MVT compares a function’s change with its derivative somewhere in between.
Apply the theorem to on , and bound using .
Theory CLP-1 §2.13Key results
Worked solution
Strategy
The MVT expresses the change in g over [0, x] as x times a derivative value at an interior point. Because the derivative decreases, that value is below its value at 0, which gives the inequality.
Working
g is continuous on [0, x] and differentiable on (0, x) for any x > 0.
Mean Value Theorem.
x is positive, so dividing by a larger positive number gives a smaller result.
Rearrange.
Answer
The answer is the inequality holds for every positive x.
What earns the marks
- Applies the MVT to g(t) = √(1 + t) on [0, x], obtaining √(1 + x) − 1 = x · g′(c) = x / (2√(1 + c)) for some c in (0, x).
- Uses c > 0 to bound the derivative: 1/(2√(1 + c)) < 1/2.
- Concludes √(1 + x) − 1 < x/2 and rearranges to the required inequality.
Check
Squaring gives a quick confirmation: (1 + x/2)² = 1 + x + x²/4, which exceeds 1 + x for every x > 0.
Where this goes wrong
Squaring both sides and checking the algebra. That verifies the inequality but does not “use the Mean Value Theorem”, which is what the question asks for and what the marks are attached to.
Leans on
Derivative of a square root, from The chain rule and logarithms
Try a similar question
Let . Show that there is a real number with .Saved in this browser only.