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

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.]
