Function that differentiates once not twice at x=0

[zymap]

Hi, I can differentiate this twice but don't understand the x=0 part. Any help is appreciated.

 

[Dr.Peterson]

Please show what you got for the derivatives, and how you determined whether they exist at x=0.

 

[Steven G]

What would have to be true if x if f is differentiable at x=0? Hint: the definition of a derivative is a limit and a limit exists when .....

 

[zymap]

Please show what you got for the derivatives, and how you determined whether they exist at x=0.

I differentiated the funtion twice but plugging x=0 in the funtions would be invalid.

 

[Steven G]

Most people here will not open up your scan. Can you just post it here?

 

[Steven G]

Why would plugging in 0 be a problem? You do realize that what you just posted goes against the answer you were given?
You say that you do not understand the x=0 part. What don't you understand? If you do not try then we really can't help you. I asked you what makes a limit exist? A limit exists if the following conditions are met. These conditions are ....
Please try to answer my questions. Just giving you answers does not work. You thinking and better understanding definitions does work?

 

[zymap]

Because I thought plugging in 0 to a function with fractions with x as the denominator is invalid

 

[zymap]

Why would plugging in 0 be a problem? You do realize that what you just posted goes against the answer you were given?
You say that you do not understand the x=0 part. What don't you understand? If you do not try then we really can't help you. I asked you what makes a limit exist? A limit exists if the following conditions are met. These conditions are ....
Please try to answer my questions. Just giving you answers does not work. You thinking and better understanding definitions does work?

a limit exists when the function is going towards a particular value

 

[HallsofIvy]

I differentiated the funtion twice but plugging x=0 in the funtions would be invalid.

Well, that's a problem. Differentiating the function where x is not 0 then evaluating that at x= 0 does not necessarily give the derivative at x= 0.

For x not 0, \(\displaystyle f(x)= x^3 sin(1/x^2)\) so \(\displaystyle f'(x)= 3x^2 sin(1/x^2)- x^3 cos(1/x^2)(-2/x^3)\)\(\displaystyle = 3x^2 sin(1/x^2)+ 2cos(1/x^2)\).

Yes, that is not defined at x= 0 but, since that is the derivative only for h not x, it does not tell us anything about the derivative AT x= 0.

For h not 0, \(\displaystyle f(h)= h^3 sin(1/h^2)\) while f(0)= 0 so \(\displaystyle \frac{f(h)- f(0)}{h}= \frac{h^3 sin(1/h^2)}{h}= h^2 sin(1/h^2)\)

To take the limit as h goes to 0, write it as \(\displaystyle \frac{sin(1/h^2)}{1/h^2}\) and let \(\displaystyle u= 1/h^2\). As h goes to 0, u goes to infinity so this limit is \(\displaystyle \lim_{u\to \infty} \frac{sin(u)}{u}\). As u goes to infinity, sin(u) always lies between -1 and 1 while the denominator, u, goes to infinity so this limit is 0.

The function is differentiable at x= 0 and the derivative there is 0.

Since the limit, as x goes to 0, of f' is not 0 (does not exist) f' is not continuous at 0 so can't be differentiable at x= 0. Therefore f itself is not twice differentiable at x= 0.

 

[zymap]

Well, that's a problem. Differentiating the function where x is not 0 then evaluating that at x= 0 does not necessarily give the derivative at x= 0.

For x not 0, \(\displaystyle f(x)= x^3 sin(1/x^2)\) so \(\displaystyle f'(x)= 3x^2 sin(1/x^2)- x^3 cos(1/x^2)(-2/x^3)\)\(\displaystyle = 3x^2 sin(1/x^2)+ 2cos(1/x^2)\).

Yes, that is not defined at x= 0 but, since that is the derivative only for h not x, it does not tell us anything about the derivative AT x= 0.

For h not 0, \(\displaystyle f(h)= h^3 sin(1/h^2)\) while f(0)= 0 so \(\displaystyle \frac{f(h)- f(0)}{h}= \frac{h^3 sin(1/h^2)}{h}= h^2 sin(1/h^2)\)

To take the limit as h goes to 0, write it as \(\displaystyle \frac{sin(1/h^2)}{1/h^2}\) and let \(\displaystyle u= 1/h^2\). As h goes to 0, u goes to infinity so this limit is \(\displaystyle \lim_{u\to \infty} \frac{sin(u)}{u}\). As u goes to infinity, sin(u) always lies between -1 and 1 while the denominator, u, goes to infinity so this limit is 0.

The function is differentiable at x= 0 and the derivative there is 0.

Since the limit, as x goes to 0, of f' is not 0 (does not exist) f' is not continuous at 0 so can't be differentiable at x= 0. Therefore f itself is not twice differentiable at x= 0.

Thank you this is a great help

 

[Steven G]

Thank you this is a great help

Yes it was helpful since it used the definition of the derivative which I suggested that you look at twice. Now that it was spoon fed to you I hope that you can remember it.