- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
- Theory
- CLP-1 §2.13Key results
The Mean Value Theorem practice question 6
How many real roots does the function below have?
Counting roots takes two arguments: one that roots exist, and one that caps how many there can be. Which theorem gives each?
For existence, look for a sign change: compare and , then use the Intermediate Value Theorem.
For the cap, compute and show it is always positive. If had two roots, what would Rolle’s theorem say about between them?
A whole number.
Theory CLP-1 §2.13Key results
Worked solution
Strategy
Existence from a sign change and the IVT; uniqueness because the derivative never vanishes, so Rolle’s theorem forbids two roots.
Working
sin 2 ≈ 0.91, so g(−1) ≈ −2.09. g is continuous, as a sum of continuous functions, so this sign change and the IVT give at least one root in (−1, 0).
The derivative is bounded away from zero.
g is differentiable everywhere. If g had two roots a < b, Rolle’s theorem would give a c between them with g′(c) = 0, which is impossible. So there is at most one root, hence exactly one.
Answer
The answer is exactly one.
Check
The line 4x + 1, rising at slope 4, meets the wave sin(2x), whose slope is at most 2, only once, near x = −0.444.
Where this goes wrong
Guessing that the oscillating sine term creates several crossings. The slope of 4x dominates: the derivative never drops below 2, so the graph only ever rises.
Leans on
The Intermediate Value Theorem, from Continuity
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