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
 A: hints:
1) Both $\sin$ and $\cos $ share the same period. Use that to show the function is not one to one. 
2) Each of $\sin $ and $\cos$ is bounded between $-1$ and $1$. Use that to show that the function is not onto by showing it is bounded between $0$ and $2$ (assuming you take as the codomain $\mathbb R$, otherwise, the question has no meaning). It appears you edited the question so the domain is now $[0,2]$. So, now to show the function is not onto, show that $0$ is never attained by remembering basic facts about the trigonometric functions. 
3) Apply the limit definition of the derivative at $0$, and show the limit does not exist. Alternatively, if the function was differentiable at $0$, then in a small enough neighborhood of $0$ the function $f(x)-\cos x$ would also be differentiable (as the difference of differentiable functions). However, that function is $|\sin x|$, and you can show that function is not differentiable at $0$ by applying the definition again. 
A: (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.
A: To see it isn't onto $[0,2]$, you only need show it is never $0$. Note that $|a|+|b|=0$ iff $a=b=0.$ For your function, this means are there any $x$ for which both $\cos x=0$ and simultaneously $\sin x=0.$ It's easy to see not, just noting that e.g. $\sin x=0$ iff $x=n \pi,$ and at those points $\cos x = \pm 1.$
Added note: The function can be viewed as the "taxicab" distance between the point $(\cos x, \sin x)$ on the unit circle to the origin $(0,0).$ In this way it seems almost clear that the minimum is $1$ and the maximum is $\sqrt{2}$ attained respectively in quadrant I when $x=0$ and when $x=\pi/4.$ [By symmetry one can restrict to quadrant I, where each of the cosine and sine are nonnegative, so the absolute value signs may be dropped and usual calc methods used for max and min.]
A: HINT: 1. $1-1$: It is periodic, but not constant. 2. (differentiability) What are one-sided derivatives at 0? 3. (not onto) $\sin^2x+\cos^2x=1$.
A: Here I plot the graph of your function in $x\in[-6\pi,6\pi].$ This is not an answer. But you can get a clear idea than intuition. 

A: hint: $f$ is $2\pi$-periodic so you only need to worry about on $[0, 2\pi]$.  you can write out expressions for both $f$ and its derivative on the four pieces $[0,\pi/2], [\pi/2, \pi],[\pi, 3\pi/2]$ and $3\pi/2, 2pi].$ in each of these interval $f = \pm \sin x \pm \cos x$ check the derivative at the boundaries $0, \pi/2,\pi, 3\pi/2$ u
