Show $2x - 1 - \sin (x)$ has exactly one real root.

Question

I have the following solution to the above problem:

Let $f(x)=2x-1-\sin(x)$

Observe, $2x-1, \sin(x)$ are continuous functions, hence by arithmetic of continuous functions $f(x)$ is continuous $\forall x$.

$f'(x)=2-\cos(x)$, hence $f'(x)>0$, $\forall x$ and hence $f(x)$ is increasing $\forall x$.

Observe $f(0)=-1<0$ and $f(\pi)=2\pi-1>0$ hence since $f(x)$ is continuous and increasing $\forall x$ there is exactly one real root.

I'm wondering if my solution would be considered acceptable and if there is an alternative way to go about solving this problem. Thanks!

Answer 1

Υοur solution is perfect.

As a slightly different proof, you can prove that $f$ has at least one root, and if you assume that has more than one roots, then by Rolle's s theorem, would exist $y\in \Bbb{R}$ such that $f'(y)=0$

Thus $\cos{y}=2$ which is a contradiction.

Answer 2

Your solution is not perfect. A function is not an expression. You can say that $2x-1,\sin(x)$ are continuous in $x∈\mathbb{R}$, but they are not functions. And $f(x)$ is not a function either. You should also have defined the domain of $f$. You can either say that $f$ is continuous on its entire domain (presumably $\mathbb{R}$), in which case you can just say "$f$ is continuous", or you can say that $f$ is continuous at every $x∈\mathbb{R}$. Do not use "$∀$" in the manner you are using it, because it is wrong. Either write the intended meaning in proper words, or use the proper syntax of first-order logic if you still want to use the symbolic form. These errors aside, your idea of using IVT is correct.