Show the Equation $2x-1-sinx=0$ has Exactly One Real Root

Question

Question :

Use the Intermediate Value Theorem and Mean Value Theorem to show that the queation $2x-1-sinx=0$ has exactly one root.

My answer :

Since we cannot compute the $y$ when $x=0$, we use the Intermediate Value Theorem.

Randomly choose two points :
When x = -10, 2(-10) -1 -sin(-10) =  -21.5, (-10,  -21.5)
When x = 100, 2(100) -1 -sin(100) = 199.51, (100, 199.51)

Since $y=2x-1 -sin(x)$ is a continuous curve and the two points exist at different sign regions, there is at least one point at $y=0$.

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For the Mean Value Theorem, how to use it for the proof?

Thank you for your help.

Answer 1

If $f(x) = 2x - 1 - \sin x$ had two distinct roots $x_1$ and $x_2$, then

$$\frac{f(x_1) - f(x_2)}{x_1 - x_2} = 0$$

Now use the mean value theorem to write this difference quotient with a derivative; there's a quick contradiction.

Answer 2

Setting

$y(x) = 2x - 1 - \sin x, \tag{1}$

we see that

$y(0) = -1 < 0 \tag{2}$

and

$y(\dfrac{\pi}{2}) = \pi - 1 - \sin \dfrac{\pi}{2} = \pi - 2 > 0, \tag{3}$

so the IVT shows there is at least one root 'twixt $0$ and $\pi / 2$. Now suppose $y(x)$ had more than one zero, say $y(\alpha) = y(\beta) = 0$ with $\alpha < \beta$. We see that

$y'(x) = 2 - \cos x > 0 \tag{3}$

for all $x \in \Bbb R$; but by the MVT there exists a point $\gamma \in (\alpha, \beta)$ with

$y'(\gamma) = \dfrac{y(\beta) - y(\alpha)}{\beta - \alpha} = 0; \tag{4}$

but this contradicts (3); thus $y(x)$ has exactly one zero in $\Bbb R$. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!