Piecewise Continuous and piecewise smooth functions

[comfortablynumb]

I do not know to start. Here is the problem.Determine if the given function is piecewise continuous, piecewise smooth, or neither. Here $x\neq0$ is in the interval $[-1,1]$ and $f(0)=0$ in all cases.

1. $f(x)=sin(\frac{1}{x})$
2. $f(x)=xsin(\frac{1}{x})$
3. $f(x)={x}^{2}sin(\frac{1}{x})$
4. $f(x)={x}^{3}sin(\frac{1}{x})$ .

 

[topsquark]

I do not know to start. Here is the problem.Determine if the given function is piecewise continuous, piecewise smooth, or neither. Here $x\neq0$ is in the interval $[-1,1]$ and $f(0)=0$ in all cases.

1. $f(x)=sin(\frac{1}{x})$
2. $f(x)=xsin(\frac{1}{x})$
3. $f(x)={x}^{2}sin(\frac{1}{x})$
4. $f(x)={x}^{3}sin(\frac{1}{x})$ .

Surely you can do the piecewise continuous part? It's just matching up if the curves are continous. So, for example, sin(1/x) has to have two limits: \(\displaystyle \lim_{x \to 0^+} f(x) = 0\) (since f(0) = 0) and \(\displaystyle \lim_{x \to 0^-} f(x) = 0\).

Piecewise smooth would be if the first derivatives are continuous. Do you need help with that part?

-Dan

 

[math771]

Hi there,

Based on the given information, it seems like the function is piecewise continuous since it is defined for all values of x except for x=0, which is a removable discontinuity. However, it is not piecewise smooth since the derivative of the function is not continuous at x=0. The function also does not have any other piecewise definitions, so it cannot be classified as neither. Hope this helps!