Is the Function h(x) Differentiable at x=0?

[kathrynag]

Homework Statement

Define h(x)=x^3sin(1/x) for x$$\neq$$0. and h(0)=0. Show h is differentiable everywhere and that h is cont everywhere, but fails to have a derivative at one point.

Homework Equations

The Attempt at a Solution

[h(x)-h(0)]/[x-0]=x^2sin(1/x)
h is diff everywhere because the limit exists and we know x$$\equiv$$0.
h'(x)=3x^2sin(1/x)-xcos(1/x)
We know f is diff at $$x_{0}$$, then f must be continuous at $$x_{0}$$.
h has a derivative at x if x$$\neq$$0.

 

[kathrynag]

I'm just unsure how to deal with it at x=0

 

[kathrynag]

Ok I see my problem. I just don't understand why we can't have h(x) with x=0. Is it because the limit wouldn't exist?

 

[jdz86]

i had a similar problem that i figured out. maybe you can apply it to yours.

Problem: Let f(x)=x^2sin1/x if x does not = 0, and f(0)=0. Find f '(x) and show that the lim as x goes to 0 of f '(x) does not exist.

My answer: f '(x)= 2xsin1/x - cos1/x
Now the lim as x goes to 0 of xsin1/x = 0, but the lim as x goes to 0 of cos1/x DNE. Since the value oscillates between -1 and 1 as x gets smaller.

~Sorry didn't have a chance to look into yours, but knew i had one similar. hope it helps.

 

[HallsofIvy]

Homework Statement

Define h(x)=x^3sin(1/x) for x$$\neq$$0. and h(0)=0. Show h is differentiable everywhere and that h is cont everywhere, but fails to have a derivative at one point.

Surely you meant "that h' is cont everywhere, but fails to have a derivative at one point."

Homework Equations

The Attempt at a Solution

[h(x)-h(0)]/[x-0]=x^2sin(1/x)
h is diff everywhere because the limit exists and we know x$$\equiv$$0.
h'(x)=3x^2sin(1/x)-xcos(1/x)
We know f is diff at $$x_{0}$$, then f must be continuous at $$x_{0}$$.
h has a derivative at x if x$$\neq$$0.

 

[HallsofIvy]

Ok I see my problem. I just don't understand why we can't have h(x) with x=0. Is it because the limit wouldn't exist?

?? h(0) is defined- it is 0. The reason for the requirement that $x\ne 0$ in the first formula is that, of course, 1/x does not exist at x= 0 so x²sin(1/x) does not exist at x= 0.

 

[samdunhamss]

The graph approaches zero but is undefined at that point there it isn't continuous at that point and shouldn't be differentiable right? I don't know

 

[HallsofIvy]

The graph approaches zero but is undefined at that point there it isn't continuous at that point and shouldn't be differentiable right? I don't know

No, the function is NOT "undefined" at x= 0. f(0)= 0. The function is continuous at 0 because $\lim_{x\rightarrow 0} f(x)= \lim_{x\rightarrow 0} x^3sin(1/x)= 0= f(0)$