Second Derivative of a Function With Absolute Values

Date: 02/27/98 at 17:37:58
From: David Smith
Subject: derivative

Dr. Math,

Let f(x) = |x|^3.  So, f(x) is the absolute value of x to the power 3. 
Find f"(0). The anwer is f"(0) = 0.

At first, I tried

   f(x) = (|x|)(|x|^2).

Then,

   f'(x) = (|x|)'(|x|^2) + (|x|)(|x|^2)'

But we know that f(x) is not differentiable at 0.
  
Also, I tried the power rule:

   f'(x) = 3(|x|)^2*(|x|)'

I got stuck at the same point too. How can I get f"(0) = 0?

Thanks

Date: 02/28/98 at 21:51:57
From: Doctor Sam
Subject: Re: derivative

David,

I usually begin problems with absolute value by rewriting the function 
in terms of its pieces. When x >= 0, |x| is just x.  When x < 0, |x| 
is its opposite, or -x.  So

                     /   x^3    when x >= 0
    f(x) = |x|^3   = |     
                     \  (-x)^3  when x < 0

Now to find the derivative:

            /   3x^2   when x > 0
    f'(x) = |     ?    when x = 0
            \  -3x^2   when x < 0

The only question is, "What happens at x = 0?" Does f(x) have a 
derivative there? The limit of (3x^2) as x approaches zero "from 
above" is 0. Likewise, the limit of (-3x^2) as x approaches zero from 
below is 0. So this function has a derivative at x = 0, and it is 0.  
So we can rewrite the derivative:

            /   3x^2   when x >= 0
    f'(x) = |   
            \  -3x^2   when x < 0

Now do the same thing to find the second derivative.

            /   6x   when x > 0
    f"(x) = |    ?   when x = 0
            \  -6x   when x < 0

As x approaches zero, both 6x and -6x approach zero, so f"(0) = 0.

-Doctor Sam, The Math Forum
 http://mathforum.org/dr.math/