calculus - To show that $f (x) = | cos x | + |sin x |$ is not one one and onto and not differentiable

Let $f : \mathbb{R} \longrightarrow [0,2]$ be defined by $f (x) = | \cos x | + |\sin x |$. I need to show that $f$ is not one one and onto. I have only intuitive idea that $\cos x$ is even function so image of $x$ and $-x$ are same. Not one to one , but how do I properly check for other things. Hints ? Thanks

Answer

(i) $f(0) = f(2\pi) = f(4\pi) = \cdots \Rightarrow$ not one to one.

(ii) Can we find $x \in \mathbb{R}$ so that $f(x) = 0 \in \operatorname{codom}f$, i.e. does $\left| \cos x \right| + \left| \sin x \right| =0$ have any real solutions?

(iii) Does $f'(0) =\displaystyle \lim_{x \to 0} \dfrac{|\cos x| + |\sin x| - |\cos 0|-|\sin 0|}{x-0}=\lim_{x \to 0} \dfrac{|\cos x| + |\sin x| - 1}{x}$ exist?

Edit:

We have

$$\lim_{x \to 0^-} \dfrac{|\cos x| + |\sin x| - 1}{x} = \lim_{x \to 0^-} \dfrac{\cos x - \sin x - 1}{x} \stackrel{\mathcal{L}}{=} \lim_{x \to 0^-} -\sin x - \cos\ x = -1.$$

But,

$$\lim_{x \to 0^+} \dfrac{|\cos x| + |\sin x| - 1}{x} = \lim_{x \to 0^+} \dfrac{\cos x + \sin x - 1}{x} \stackrel{\mathcal{L}}{=} \lim_{x \to 0^+} -\sin x + \cos\ x = 1.$$

Therefore,

$$\lim_{x \to 0} \dfrac{|\cos x| + |\sin x| - |\cos 0|-|\sin 0|}{x-0}$$

does not exist. Hence, $f$ is not differentiable on its entire domain.