Math Help - derivative of a function

1. derivative of a function

Hi, can someone cheak my solution to these problems:

For 1 a) i got they don't exist since for a>0- 3a^2
a<0- -3a^2

b) again same they don't exist since for a>0- 2a-1
a<0- -2a+1

Thsnks

Ps i put it in this section as i am learing this in an analysis course.

2. Originally Posted by SubZero
Hi, can someone cheak my solution to these problems:

For 1 a) i got they don't exist since for a>0- 3a^2
a<0- -3a^2

b) again same they don't exist since for a>0- 2a-1
a<0- -2a+1

Thsnks

Ps i put it in this section as i am learing this in an analysis course.
$\lim_{x\to0}\frac{x^2\left| x\right|-0}{x}=\lim_{x\to0}x|x|=0$

3. Originally Posted by Drexel28
$\lim_{x\to0}\frac{x^2\left| x\right|-0}{x}=\lim_{x\to0}x|x|=0$
Hi Drexel28, I thought

|x| is not differentiable at 0 since |X|/ X =

1 if X >0
-1 if X<0

4. I suggest that you graph both of these.
I think that you will be surprise by one of them.
Of course, both have derivatives everywhere except possibly at one point.

5. Originally Posted by SubZero
Hi Drexel28, I thought

|x| is not differentiable at 0 since |X|/ X =

1 if X >0
-1 if X<0
We have that $f(x)=x^2|x|$, right? So, to claim that $f'(0)$ exists is to claim that $\lim_{x\to0}\frac{f(x)-f(0)}{x-0.}=\lim_{x\to0}\frac{x^2|x|}{x}=\lim_{x\to0}x|x|$ exists. But, $-x^2\leqslant x|x|\leqslant x^2$ where it follows from the squeeze theorem that $\lim_{x\to0}x|x|=0=f'(0)$.