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Analysis and the Derivative


Date: 01/23/2001 at 14:19:58
From: Yvonne
Subject: Analysis and the Derivative

In my analysis class, we have come to the chapter on the derivative. 
The text suggests considering the following problem. I'm having a 
little trouble starting and I would greatly appreciate it if you could 
help me a little.

Suppose that f:RtoR is differentiable at c and that f(c)=0.  Show 
that g(x):=|f(x)| is differentiable at c if and only if f'(c)=0.  

Any suggestions would be terrific. Thanks.


Date: 01/24/2001 at 21:36:14
From: Doctor Fenton
Subject: Re: Analysis and the Derivative

Dear Yvonne,

Thanks for writing to Dr. Math. For this problem, you need to use the 
definition of derivative:

  a function F(x) is differentiable at x = a if

         F(x)-F(a)
     lim ---------  exists.  The value of this limit is F'(a).
    x->a   x - a

So, you know that (1) f(c)=0 ;
and               (2) f is differentiable at x = c, or in other words,

         f(x)-f(c)
     lim ---------  = f'(c).
    x->c   x - c

I think you'll need the epsilon-delta version of this: (I'll write e
for epsilon and d for delta)

  Given e>0, there is a d>0 such that for all x such that

  0<|x-c|<d, | f(x)-f(c)         |
             | ---------- - f'(c)| < e  .
             |   x - c           |

To show g(x) is differentiable at x = c, you must show that

         g(x)-g(c)
     lim ---------  exists.
    x->c   x - c

That is, you must find a number L such that, given any e > 0, you can
find a d>0 such that if x satisfies

0 < |x-c| < d, then

             | g(x)-g(c)      |
             | ---------- - L | < e  .
             |   x - c        |

If you look at a function near a point where it crosses the axis, and 
if the function has a non-zero slope there (so f'(c) =/= 0, where
=/= means "is not equal to"), then the graph will look like

                                                     \   /
            /                                         \ /
     ------/----- ,  so g(x) = |f(x)| looks like  ---------- ,
          /

because taking the absolute value reflects or folds the part of the 
graph below the x-axis up into the part above the axis. That 
intuitively explains why f'(c) must = 0 in order for |f(x)| to be
differentiable at c. You should be able to see that, in that case,
the derivative of g at c, g'(c), should also be 0, so we expect L=0.

Since f(c)=0, and if we assume that f'(c)=0, then we know that 
(putting in 0 for f(c) and f'(c) in the e-d definition above,

  Given e>0, there is a d>0 such that for all x such that

  0<|x-c|<d, |    f(x)    |
             | ---------- | < e  .
             |   x - c    |

We must consider 

    | g(x)-g(c)        |
    |---------- - (0)  |
    |  x - c           |   , since as I indicated, we expect g'(c)=0 .

But 
     g(c)=|f(c)|

         = 0 ,

so this is

    | |f(x)|  |   | f(x)  |
    |-------- | = |-------| 
    |  x - c  |   | x - c |  ,

but this is exactly the expression above which we know is < e. 
So the d you chose for f also works for g.

For the converse, it's easier to use contrapositive reasoning.  
Assume that f'(c) =/= 0 .

Then the two-sided limit

         f(x)-f(c)
     lim ---------  = f'(c) exists, 
    x->c   x - c

so in particular, the two one-sided limits 

          f(x)-f(c)                    f(x)-f(c)
     lim  ---------  = f'(c) , and lim --------- = f'(c) ,
    x->c+   x - c                 x->c-  x - c

or

             f(x)                        f(x)
     lim  ---------  = f'(c) , and lim --------- = f'(c) ,
    x->c+   x - c                 x->c-  x - c

since f(c) = 0.  But then (remember that g(c)=0)


             g(x)           |f(x)|
     lim  ---------  = lim --------- ,
    x->c+   x - c     x->c+  x - c

and

             g(x)           |f(x)|
     lim  ---------  = lim --------- .
    x->c-   x - c     x->c-  x - c

But if f'(c) =/= 0, then it is either positive or negative. Assume it 
is positive.  Then

         f(x)-f(c)
     lim ---------  = f'(c)  
    x->c+  x - c
                    > 0 ,

so all the quotients f(x)/(x-c) must be positive, for x close to c.  
The denominator is clearly positive, so the numerator must also be 
positive, and f(x)>0 for x>c (i.e. in some small interval c<x<c+d).  
Similarly,

         f(x)-f(c)
     lim ---------  = f'(c) 
    x->c-  x - c
                    > 0 ,

and now the denominators are all negative, so the numerator must also 
be negative, for x to the left of c.

But g(x)=|f(x)| is either f(x) (if f(x) is positive), or g(x) is -f(x)
(if f(x) is negative).

             g(x)           |f(x)|            f(x)
     lim  ---------  = lim --------- = lim ---------- = f'(c),
    x->c+   x - c     x->c+  x - c    x->c+  x - c

and

             g(x)           |f(x)|           - f(x)
     lim  ---------  = lim --------- = lim  --------- = - f'(c).
    x->c-   x - c     x->c-  x - c     x->c-  x - c

If f'(c) =/= 0, the two one-sided limits are unequal, so the two-sided
limit does not exist.

If you have further questions, please write again.

- Doctor Fenton, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Analysis
High School Analysis

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