Injective Functions – Proving One-to-One Property

Question

The question is to show that the function $ f(x) = 2x + |\cos x|$ is both one-one and onto. I have managed to show that it is onto but got stuck at proving that it is also one-to-one.

To prove this I thought of using the first derivative and showing that the function is always increasing. For both cases of the absolute value, the derivative is $f'(x) = 2 – \sin x$ or $f'(x)=2+ \sin x$ which are both strictly greater than zero for all real values of x. However, at the points where $\cos x$ changes sign, the function is not differentiable. My question is, how can I show that the function is increasing at these points as well? Is there some calculus-based method for these points as well?

Answer 1

Here, "increasing" is understood as strictly increasing.

If $f$ is increasing on $(a,b)$ and on $(b,c)$ and continuous at $b$, then $f$ is increasing on $(a,c)$.

More details:

• If $f$ is increasing on $(a,b)$ and left continuous at $b$, then $f$ is increasing on $(a,b]$, since $$\forall x\in(a,b)\quad f(x)<f\left(\frac{x+b}2\right)\le\lim_{b^-}f=f(b).$$
• Similarly, if $f$ is increasing on $(b,c)$ and right continuous at $b$, then $f$ is increasing on $[b,c)$.
• If $f$ is increasing on $(a,b]$ and on $[b,c)$, then it increasing on $(a,c)$.