I have the function; $$ f(x) = \frac{x^2+1}{|x|+1}$$
I sketched the graph and saw that there are two local minima on it. They seem from the graph to be around $(x,y) = (-0.5, 0.825)$ and $(x,y) = (0.5, 0.825).$
I am supposed to determine all local and absolute extreme points, all asymptotes (vertical, horisontal and oblique) as well as where the function is concave and convex respectively (which, of course, has to do with the local minima and maxima according to my knowledge).
I assume I should take the derivative of the function and when $f'(x) = 0$. I can look at $f''(x)$ and where $f'(x) = 0$ and f''(x) > 0 , there is a local minimum. Where $f'(x) = 0$ and $f''(x) < 0$ , there is a local maximum. My problem is to take the derivative, I still find it a bit hard with more complex functions, but I did the quotient rule like this, is it correct?; $$f'(x) = \frac{(2x \cdot (|x| + 1)) - ((x^2+1) \cdot 1)}{(|x|+1)^2}$$
I think its a bit hard also doing the derivative of the absolute value, I'm not sure this is correct? If it is $f'(0) = -1$ and; $$ f''(x) = \frac{((2x+2)\cdot(|x|^2 +2|x| +1))-((2x+2)(2x|x|+2x-x^2+1))}{(|x|^2+2|x|+1)^2}$$
And so now I want to investigate $f''(-1)$ , right? If my long and complicated expression for $f'(x)$ and $f''(x)$ is right, which I a little bit doubt? So ; $$f'(-1) = -0.6$$
So this means a maximum is at $(-1, 0.6).$ I know from looking at the graph I should have $2$ minima at $(x, y) = (-0.5, 0.825)$ and $(x, y) = (0.5, 0.825)$ and I know the derivative's graph shows it has two asymptotes, one at $y = -1$ and one at $y = 1$. So I think I'm going wrong somewhere, can someone help me get some clarity in this? :)
The problem is that the absolute value itself is not derivable at the origin and thus there is no formula for its derivative, so you have to go back to the definition of derivative. For your function, you have: $$ f'(0)= \lim_{h\to 0}\frac{f(h)-f(0)}{h}= \lim_{h\to 0}\frac{\frac{h^2+1}{|h|+1}-1}{h}= \lim_{h\to 0}\frac{h^2-|h|}{h\,(|h|+1)} $$
Then you have to consider both cases $h>0$ and $h<0$ (i.e. lateral limits): $$ \lim_{h\to 0^+}\frac{h^2-|h|}{h\,(|h|+1)}=\lim_{h\to 0^+}\frac{h^2-h}{h\,(h+1)}=\lim_{h\to 0^+}\frac{h-1}{ h+1}=-1 $$ and $$ \lim_{h\to 0^-}\frac{h^2-|h|}{h\,(|h|+1)}=\lim_{h\to 0^-}\frac{h^2+h}{h\,(-h+1)}=\lim_{h\to 0^-}\frac{h+1}{ -h+1}=1 $$
So the derivative at $0$ does not exist and you have to think about this point through some indirect approach. For instance, if the function is always decreasing in a neighbourhood at the right of $0$ and increasing in a neighbourhood at the left of $0$, you can conclude that there is a minimum at $0$.