Expert: Paul Klarreich - 3/8/2009

Question
This is a very basic and fast question, but it has a little twist so I just wanted to be sure my thinking is correct:

Given the function f defined for all real numbers x by f(x)= 2|x-1|x^2

c.) What value of x is the derivative of f(x) continuous?

I'm not use to finding the ones contain absolute values but i think it is something like f'(x)= |x-1|2x+((x-1)/|x-1|)x^2

I think it is continuous for all real numbers except |x-1|≠ 0 right, making -1 the only undefined point because the derivative of an absolute function rule when |u|=0, in this case x-1 being u? Please correct me if I'm wrong.

Answer
Questioner:   Linrosa
Category:  Calculus
Private:  No
 
Subject:  Continuity of a derivative of an absolute value
Question:  This is a very basic and fast question, but it has a little twist so I just wanted to be sure my thinking is correct:

Given the function f defined for all real numbers x by f(x)= 2|x-1|x^2

c.) What value of x is the derivative of f(x) continuous?

I'm not use (USED TO) to finding the ones contain absolute values but i think it is something like f'(x)= |x-1|2x+((x-1)/|x-1|)x^2

I think it is continuous for all real numbers except |x-1|≠ 0 right, making -1 the only undefined point because the derivative of an absolute function rule when |u|=0, in this case x-1 being u? Please correct me if I'm wrong.

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When you deal with absolute value, the thing to do is replace it with its definition:

NOT SO FAST, BUT VERY BASIC.

               { thing, when the thing is positive
| thing | = {
               { opposite of thing, when the thing is negative
So

           { x - 1, when x-1 is positive, which means x >= 1
| x - 1 | = {
           { 1 - x, when x-1 is negative, which means x < 1

Now you can go to work:

f(x)= |x-1|x^2   << I skipped the '2', it is irrelevant.


           { (x - 1)x^2, when x-1 is positive, which means x >= 1
|x-1|x^2  = {
           { (1 - x)x^2, when x-1 is negative, which means x < 1

           { x^3 - x^2, when x-1 is positive, which means x >= 1
         = {
           { x^2 - x^3, when x-1 is negative, which means x < 1

Now you can see that it is simply two polynomials, and the only issue is "what happens at x = 1 ?"

Using the 'right side':  f'(x) = 3x^2 - 2x
Lim x->0 f'(x) = f'(0) = 0

Using the 'left  side':  f'(x) = -3x^2 + 2x
Lim x->0 f'(x) = f'(0) = 0, which is the same.

Looks like it is continuous.