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So I have this question here:

$f(x)=\frac{1}{1+|x|}+\frac{1}{1+|x-a|},a>0$

I am asked to find the derivative and I correctly found it as:

$f'(x)=-\frac{x-a}{|x-a|(|x-a|+1)^2}-\frac{x}{|x|(|x|+1)^2},a>0$

I am then asked to determine if $f'(0)$ and $f'(a)$ exist.

My understanding is that they don't exist because the derivative of $|x|$and$|x-a|$ don't exist at 0 and a by the definition of the derivative (There is a corner at those particular points.)

Is that correct? Or am I making a mistake?

Future Math person
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  • is your argument that if you naively plug in $0$ or $a$ in your formula for $f'(x)$ you get nonsense? Because this argument is invalid. – M. Van Oct 10 '16 at 17:54
  • Simplify the problem to understand better: first examine the derivative at $x = 0$ of the simpler function $$ f(x) = {1 \over 1 + |x|}. $$ Graph it, study it in a neighborhood of $x$, until you have a clear and detailed understanding of its structure. – user8960 Oct 10 '16 at 17:55
  • @Takhteh_pareh it is definded in $0$, $f(0)=1+\frac{1}{1+|a|}$. – M. Van Oct 10 '16 at 17:56
  • No I wasn't thinking about it that way M. Van. I tried to consider the behavior of the graph. – Future Math person Oct 10 '16 at 17:58
  • try to determine the limit – M. Van Oct 10 '16 at 17:59
  • User, if I try to look at the graph of the simpler function, there is a cusp at x=0. The derivative can't possibly exist there right? M. Van, if I take the limit, then the function is still well defined. It is not indeterminate or undefined anywhere. Why is that significant? – Future Math person Oct 10 '16 at 18:01
  • FURTHER EDIT: If I try to use he definition at a point, everything sill works out. Is that the right approach? – Future Math person Oct 10 '16 at 18:06

1 Answers1

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Remember that: $$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$ This approach looks more promising. Now divide the problem into simpler parts to see what's going on. For example, let $g(x)=\frac{1}{1+|x|}$ and $f(x)=g(x)+h(x)$. Then: $$\begin{align}h'(a)=\lim_{x\to a}\frac{h(x)-h(a)}{x-a}=\lim_{x\to a}\frac{\frac{1}{1+|x-a|}-1}{x-a}&=\lim_{x\to a}\frac{|x-a|}{x-a}\left(\frac{-1}{1+|x-a|}\right)\\&=-\lim_{x\to a}\frac{|x-a|}{x-a} \end{align}$$ You see that if you approach $a$ from the left you will get $+1$ and if you approach it from the right you'll get $-1$. Hence $h'(a)$ does not exist, whereas $g'(a)$ does exist. This implies $f'(a)$ does not exist. Because if it existed, $h'(a)=f'(a)-g'(a)$ and you would get a contradiction.

The same statement goes for $g'(x)$ at $x=0$.

polfosol
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