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
$\displaystyle f(x) = 7x^3 - 6$?
If so
$\displaystyle f(x + h) = 7(x + h)^2 - 6$
$\displaystyle = 7(x^2 + 2xh + h^2) - 6$
$\displaystyle = 7x^2 + 14xh + 7h^2 - 6$.
Therefore
$\displaystyle f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$
$\displaystyle = \lim_{h \to 0}\frac{7x^2 + 14xh + 7h^2 - 6 - (7x^2 - 6)}{h}$
$\displaystyle = \lim_{h \to 0}\frac{14xh + 7h^2}{h}$
$\displaystyle = \lim_{h \to 0}(14x + 7h)$
$\displaystyle = 14x$.
Oh yeah... oops...
$\displaystyle f(x) = 7x^3 - 6$.
$\displaystyle f(x + h) = 7(x + h)^3 - 6$
$\displaystyle = 7(x^3 + 3x^2h + 3xh^2 + h^3) - 6$
$\displaystyle = 7x^3 + 21x^2h + 21xh^2 + 7h^3 - 6$.
Now work out
$\displaystyle \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$.