Differentiability/continuity of polynomials. [f(x)=0 for x=0, x^2 for x!=0]

[AvgStudent]

Seeing recent calculus posts, raised this question for me.
I know that polynomials are differential everywhere, but what if I define the function like this:
[math]f(x) =\begin{cases} 0 & \text{ for } x=0\\ x^2 & \text{ for } x\neq0 \end{cases}[/math]Would we say this function is continuous everywhere, but not differentiable at x=0?

 

[Steven G]

No, we would say that your function is continuous everywhere and differentiable everywhere.

The way that you defined f(x) is exactly one way of defining f(x) = x^2.

What makes you think that it isn't differentiable at x=0?