1. Difference quotient question

Find the difference quotient of f

f(x)=[7x^3]-6

Not really sure where to go with this one, maybe cause its late I'm not thinking correctly. Any help is greatly appreciated.

2. Originally Posted by Khaotical
Find the difference quotient of f

f(x)=[7x^3]-6

Not really sure where to go with this one, maybe cause its late I'm not thinking correctly. Any help is greatly appreciated.
Is it

$f(x) = 7x^3 - 6$?

If so

$f(x + h) = 7(x + h)^2 - 6$

$= 7(x^2 + 2xh + h^2) - 6$

$= 7x^2 + 14xh + 7h^2 - 6$.

Therefore

$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$

$= \lim_{h \to 0}\frac{7x^2 + 14xh + 7h^2 - 6 - (7x^2 - 6)}{h}$

$= \lim_{h \to 0}\frac{14xh + 7h^2}{h}$

$= \lim_{h \to 0}(14x + 7h)$

$= 14x$.

3. Originally Posted by Prove It
Is it

$f(x) = 7x^3 - 6$?

If so

$f(x + h) = 7(x + h)^2 - 6$

$= 7(x^2 + 2xh + h^2) - 6$

$= 7x^2 + 14xh + 7h^2 - 6$.

Therefore

$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$

$= \lim_{h \to 0}\frac{7x^2 + 14xh + 7h^2 - 6 - (7x^2 - 6)}{h}$

$= \lim_{h \to 0}\frac{14xh + 7h^2}{h}$

$= \lim_{h \to 0}(14x + 7h)$

$= 14x$.
It's cubed, not squared, in the initial problem. But the technique is okay.

4. Originally Posted by Matt Westwood
It's cubed, not squared, in the initial problem. But the technique is okay.
Oh yeah... oops...

$f(x) = 7x^3 - 6$.

$f(x + h) = 7(x + h)^3 - 6$

$= 7(x^3 + 3x^2h + 3xh^2 + h^3) - 6$

$= 7x^3 + 21x^2h + 21xh^2 + 7h^3 - 6$.

Now work out

$\lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$.