Show it is differentiable and compute f '

thaopanda · #1 $Old$ November 1st, 2009, 07:59 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 · #2 $Old$ November 1st, 2009, 08:07 PM

Quote:

Originally Posted by thaopanda $View Post$

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 · #3 $Old$ November 1st, 2009, 08:57 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 · #4 $Old$ November 1st, 2009, 11:30 PM

Quote:

Originally Posted by thaopanda $View Post$

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 $\lim_{x \to 0^{+}}x^2\sin{\left(\frac{1}{x}\right)} = 0$ .