Let $f(x)=\sin x-2x(x-1)$ then we need to find two roots $\alpha$ and $\beta$ such that $\alpha<\beta$ Then i think to apply $IVT$ based on the following conditions
1)$f$ is strictly monotone in the domain $D$
2)$f(\alpha)<0$ and $f(\beta)>0$ then $f(\alpha)\times f(\beta)<0$
Now my question how to prove $f$ is strictly monotone if $f'(x)=\cos x-4x+2$
Is there any short cut to prove this using analysis only?
On
Note that $x=0$ is the trivial root. Then, it remains to prove that
$$f(x)= 2(x-1)-\frac{\sin x}x $$
have exactly one root. Evaluate
$$f’(x) = \frac{2x^2-x\cos x+\sin x}{x^2} =\frac1{x^2} \int_0^x t(4+\sin t)dt>0 $$
for all $x\in R$. That is $f(x)$ monotonically increases, along with $f(\pm\infty)=\pm\infty$. Thus, $f(x)$ crosses the $x$-axis only once, hence exactly one root.
If it were true that $f$ is strictly monotone in $\mathbb{R}$, then $f$ would have no more than $1$ real zero. So, that's false. Notice that $f''(x)=-\sin(x)-4<0$, hence $f'$ is strictly decreasing in $\mathbb{R}$. Since $f'(0)=3>0$ and $f'(1)=\cos(1)-2<0$, $f'$ has a unique zero $x_0$, and $x_0\in(0,1)$. Thus, $f$ is strictly monotone in each of the intervals $(-\infty,x_0]$ and $[x_0,\infty)$ where it can have no more than one zero. So, $f$ has at most $2$ real zeros. Now, you have to show $f$ has at least $2$ zeros to conclude. You can check that $f(0)=0,f(1)>0,f(2)<0$, and apply the IVT for the interval $[1,2]$.