more differntiable

• Apr 22nd 2007, 12:36 PM
learn18
more differntiable
let f(X) = x^2 sin(1/x) for x is not equal to 0 and f(0)= 0

A) use the chain rule and the product rule to show that f is differentiable at each c is not equal to 0 and find f'(c)
B) show that f is differentiable at x = 0 and show that f(0)=0
C) show that f' is not continuous at x = 0
• Apr 22nd 2007, 01:10 PM
Jhevon
Quote:

Originally Posted by learn18
let f(X) = x^2 sin(1/x) for x is not equal to 0 and f(0)= 0

A) use the chain rule and the product rule to show that f is differentiable at each c is not equal to 0 and find f'(c)

at x = c not= 0, f(x) is continuous and hence differentiable.

so f(c) = (c^2)sin(1/c)
=> f ' (c) = 2csin(1/c) + (-c^-2)(c^2)cos(1/c)
=> f ' (c) = 2csin(1/c) - cos(1/c)

Quote:

B) show that f is differentiable at x = 0 and show that f(0)=0
we must show that f is continuous at x = 0

note that |(x^2)sin(1/x)|<or= x^2 for all x

so we have: -x^2 <or= (x^2)sin(1/x) <or= x^2
=> lim{x-->0+} -x^2 <or= lim{x-->0+}(x^2)sin(1/x) <or= lim{x-->0+}x^2
=> 0 <or= lim{x-->0+}(x^2)sin(1/x) <or= 0
so by the squeeze theorem:

lim{x-->0+}(x^2)sin(1/x) = 0
similarly, we can show, lim{x-->0-}(x^2)sin(1/x) = 0

thus f is continuos at x = 0, so by the definition of continuity we have f(0) = lim{x-->0}(x^2)sin(1/x) = 0

Quote:

C) show that f' is not continuous at x = 0
this seems trivial to me, but i may be wrong. at x = 0, 1/x is undefined, so f ' (x) = 2xsin(1/x) - cos(1/x) is undefined and is not continuous. you can take the left and right hand limits to see this, show that they are different or something
• Apr 22nd 2007, 01:32 PM
ThePerfectHacker
Quote:

Originally Posted by Jhevon

note that |(x^2)sin(1/x)|<or= x^2 for all x

so we have: -x^2 <or= (x^2)sin(1/x) <or= x^2
=> lim{x-->0+} -x^2 <or= lim{x-->0+}(x^2)sin(1/x) <or= lim{x-->0+}x^2
=> 0 <or= lim{x-->0+}(x^2)sin(1/x) <or= 0
so by the squeeze theorem:

lim{x-->0+}(x^2)sin(1/x) = 0
similarly, we can show, lim{x-->0-}(x^2)sin(1/x) = 0

thus f is continuos at x = 0, so by the definition of continuity we have f(0) = lim{x-->0}(x^2)sin(1/x) = 0

But this shows that f is continous at 0 not differenciable at 0
• Apr 22nd 2007, 01:41 PM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
But this shows that f is continous at 0 not differenciable at 0

there are three things that can cause a function f to be non-differentiable, is there not?

if:

(1) f is undefined
(2) f is discontinuous
(3) f has a sharp turn

case 3 doesn't happen, and case (1) doesn't happen because of the way the function is defined (by definition, f(0) = 0, so it's not undefined there as the problem is phrased), so i thought i could just take care of case (3) and be done with it, which is what i did

and i forgot to say "this is my 13:):)th post!"
darn it, i missed it again!
• Apr 22nd 2007, 01:46 PM
ThePerfectHacker
Quote:

Originally Posted by learn18
let f(X) = x^2 sin(1/x) for x is not equal to 0 and f(0)= 0

B) show that f is differentiable at x = 0 and show that f(0)=0

First f(x) is defined on an open interval containing 0.
Thus we can consider the derivative limit:

lim(x-->0) (f(x) -f(0))/(x-0) = lim(x-->0) f(x)/x

lim(x-->0) x^2 sin(1/x)/x = lim (x-->0) x*sin(1/x)

The limit exists because:

-1<=sin(1/x) <= 1 then

-x<= x*sin(1/x) <= x for x>0
x<= x*sin (1/x) <= -x for x<0
In both cases the limit as x-->0 is 0 by Squeeze Theorem.

Thus, f(x) is differenciable at 0, and furthermore the derivative is 0.