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

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. 
