## anonymous 3 years ago find f ' (x) and f '(c) f(x)= (1/3)(2x^3 - 4) value of c = 0

1. anonymous

i'm not really sure how to get started

2. across

$f(x)=\frac13\left(2x^3-4\right)\\ f'(x)=2x^2\\ f'(0)=0.$

3. anonymous

how did you get those primes?

4. across

I used both$f(x)=ax^{b}\Longrightarrow f'(x)=abx^{b-1},$and$f(x)=c\Longrightarrow f'(x)=0.$

5. anonymous

ah ok, thanks