Show it is differentiable and compute f '

**thaopanda** · November 1st 2009, 05:59 PM

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

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

is differentiable and compute f '.

**Prove It** · November 1st 2009, 06:07 PM

Originally Posted by thaopanda

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

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

is differentiable and compute f '.

What about when $x > 0$ ?

**thaopanda** · November 1st 2009, 06:57 PM

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

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

sorry about that

**Prove It** · November 1st 2009, 09:30 PM

Originally Posted by thaopanda

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

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

sorry about that

To be differentiable, the function needs to be continuous and smooth.

Clearly $x^3$ is continuous since it is a polynomial.

Also, since $x^2$ and $\sin{\left(\frac{1}{x}\right)}$ are both continuous for $x > 0$ , so will be their product.

So the only place where this hybrid function might not be continuous is where the function "changes" - so at $x = 0$ . 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 $\lim_{x \to 0^{+}}x^2\sin{\left(\frac{1}{x}\right)} = 0$ .

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