2
$\begingroup$

Question :

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

2x-1-sinx=0

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 $y=0$.


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

Thank you for your help.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
0
2
$\begingroup$

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!!!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.