Expert: Paul Klarreich - 12/28/2006

Question
f(x)=abs(x)+ xsin(x)

Xmin=-2Pie     Xmax=2Pie
Ymin=-5        Ymax-10

1. Prove that f(x) is an even function.
2. Is f(x) continuous at x=0? Explain.
3. Does f'(x) exist? Explain.  

Answer
Questioner:  x
Category:  Calculus
Private:  no
 
Subject:  applications of derivatives
Question:  f(x)= abs(x)+ x sin(x)

1. Prove that f(x) is an even function.
2. Is f(x) continuous at x=0? Explain.
3. Does f'(x) exist? Explain.

.........................................
Hi, x, (Is that really the name your parents gave you?)

For your function:

f(x) = abs(x) + x sin(x),

You can apply these definitions and rules:

f(x) is even if  f(-x) is the same as f(x)
The sum of two    even   functions is even.
The sum of two continuous functions is continuous.
The sum of two differentiable functions is differentiable.


A. f(-x) = abs(-x) + (-x)(sin(-x))

Now  abs(x) can be written many ways.  One of them is:

abs(x) = sqrt(x^2)

[I am sure your precalculus teacher told you every day and twice on Tuesdays, that  sqrt(x^2) was not just x.]

Then abs(-x) = sqrt( (-x)^2 ) = sqrt( x^2 ) = abs(x), and abs(x) is even.
Now sin(-x) = - sin(x), because sin(x) is an ODD function.


Back to :

f(-x) = abs(-x) + (-x)(sin(-x))

= abs(x) + (-x)(- sin(x))
= abs(x) + x sin(x)
= f(x)
proving f(x) is even.

B. abs(x) is continuous at x = 0. Another definition of abs(x) is rule-based:

abs(x) =  |  x,  when x >= 0   [Rule R]
         | -x,  when x < 0    [Rule L]

Now abs(0) = 0, applying Rule R.  What about limits?

lim  abs(x) =
x->0+

lim  x = 0, applying Rule R
x->0+

lim  abs(x) =
x->0-

lim -x = 0, applying Rule L
x->0-

These limits are the same, so lim abs(x) = 0 = abs(0), making the function continuous.

Since x and sin x are also continuous at x = 0, so is f(x).

C. Now differentiability.  Obviously the only 'issue' is what happens at x = 0?

[NOTE: see an earlier answer from me about DIFFERENTIATION OF ABSOLUTE VALUES.]

Assume  g(x) = abs(x).  What is g'(0)?  Does it exist?  Look at the left and

right-hand limits.
     g(0 + h) - g(0)
lim  --------------- =
x->0+       h

     0 + h - 0
lim  ----------- =  [Using rule R]
x->0+     h

      h
lim  ---- =
x->0+  h

lim  1 =  1  as the right-hand limit.
x->0+
. . . . . . . . . .
     g(0 + h) - g(0)
lim  --------------- =
x->0-       h

     -(0 + h) - 0
lim  ------------ =  [Using rule L]
x->0-      h

      -h
lim  ---- =
x->0-  h

lim  -1 =  -1  as the left-hand limit.
x->0-

Since these limits are not the same, the LIMIT does not exist, the derivative does not exist at x = 0, so your entire derivative f'(x) does not exist at  x = 0.  [No problem for any other x, of course.]

In the future, please provide a name.  I am sure your fingers must have slipped when you entered the name part.

On the other hand, if you left out your name because that might reveal to your teacher that you got the answer from me, that is unethical, unless you cite this source when handing in the solution.

I leave it to you to judge which happened.