# Thread: Show it is differentiable and compute f '

1. ## Show it is differentiable and compute f '

Show that the function f: R $\displaystyle \rightarrow$ R given by

f(x): =
$\displaystyle x^3, x \leq 0$
0, x = 0

is differentiable and compute f '.

2. Originally Posted by thaopanda
Show that the function f: R $\displaystyle \rightarrow$ R given by

f(x): =
$\displaystyle x^3, x \leq 0$
0, x = 0

is differentiable and compute f '.
What about when $\displaystyle x > 0$?

3. oh wow, I wasn't paying attention. Messed that one up. It's actually:

f(x) :=
$\displaystyle x^3, x \leq 0$
$\displaystyle x^2*sin(\frac{1}{x}), x > 0$

4. Originally Posted by thaopanda
oh wow, I wasn't paying attention. Messed that one up. It's actually:

f(x) :=
$\displaystyle x^3, x \leq 0$
$\displaystyle x^2*sin(\frac{1}{x}), x > 0$

Clearly $\displaystyle x^3$ is continuous since it is a polynomial.
Also, since $\displaystyle x^2$ and $\displaystyle \sin{\left(\frac{1}{x}\right)}$ are both continuous for $\displaystyle x > 0$, so will be their product.
So the only place where this hybrid function might not be continuous is where the function "changes" - so at $\displaystyle x = 0$. We need to check that the left hand and right hand limits are 0.
Clearly the left hand limit is 0. So we would need to show that $\displaystyle \lim_{x \to 0^{+}}x^2\sin{\left(\frac{1}{x}\right)} = 0$.