- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
- Theory
- CLP-1 §2.13Key results
The Mean Value Theorem practice question 5
Let . Show that there is a real number with .
Solving by algebra is hopeless here. Which theorem promises a zero of without finding it, and what does it need from ?
Rolle’s theorem: if is continuous on , differentiable on , and , then somewhere in between. Look for two inputs with equal outputs.
Try and ; the sine term vanishes at both.
Theory CLP-1 §2.13Key results
Worked solution
Strategy
Solving f′(c) = 0 directly means solving 2c − 4π + 3cos c = 0, which has no closed form. Rolle’s theorem sidesteps that: two equal values force a zero derivative in between.
Working
f is a polynomial plus 3 sin x, so it is continuous and differentiable on all of ℝ; the hypotheses hold on any closed interval.
Two inputs with the same output.
By Rolle’s theorem there is a c in (0, 4π) with f′(c) = 0.
Answer
The answer is by Rolle’s theorem on the interval from 0 to 4π, some c there has f prime of c equal to 0.
What earns the marks
- Notes that f is continuous and differentiable everywhere (a polynomial plus a multiple of sine).
- Finds two points where f takes the same value, e.g. f(0) = 0 and f(4π) = 16π² − 16π² + 3 sin(4π) = 0.
- Invokes Rolle’s theorem (or the MVT with zero secant slope) on [0, 4π] to conclude that f′(c) = 0 for some c in (0, 4π).
Check
f′(x) = 2x − 4π + 3 cos x is about −9.57 at 0 and 15.57 at 4π, so it does pass through zero in between.
Where this goes wrong
Trying to solve 2c − 4π + 3cos c = 0 explicitly. The question says “show that there is”, which is an invitation to an existence argument, not a calculation.
Leans on
Values of sine at multiples of π, from Functions, graphs and inverses
Try a similar question
Use the Mean Value Theorem to show that for every .Saved in this browser only.