Show that the slope of y = x^3 - 3x + 3 at any x is 3x^2 - 3
A=delta (dont know how to do delta)
im doing it using this method
( (x+Ax)^3 - 3(x+Ax) + 3 ) - ( x^3 - 3x + 3 ) / ( (x+Ax) - x )
i simplify and get Ax^2 + 3xAx + 3x^2 - 6x
Show that the slope of y = x^3 - 3x + 3 at any x is 3x^2 - 3
A=delta (dont know how to do delta)
im doing it using this method
( (x+Ax)^3 - 3(x+Ax) + 3 ) - ( x^3 - 3x + 3 ) / ( (x+Ax) - x )
i simplify and get Ax^2 + 3xAx + 3x^2 - 6x
where did 6x come from?
you simplified something wrong. expanding everything, you should get:
$\displaystyle \lim_{\Delta x \to 0} \frac {x^3 + 3x^2(\Delta x) + 3x(\Delta x)^2 + (\Delta x)^3 - 3x - 3 \Delta x + 3 - x^3 + 3x - 3}{\Delta x}$
now cancel what is to be canceled, simplify as much as possible, and take the limit
let $\displaystyle h = \Delta x$
$\displaystyle f(x) = x^3 - 3x + 3$
$\displaystyle f(x+h) = (x+h)^3 - 3(x+h) + 3$
$\displaystyle \lim_{h \to 0} \frac{(x+h)^3 - 3(x+h) + 3 - (x^3 - 3x + 3)}{h}$
$\displaystyle \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - 3x - 3h + 3 - (x^3 - 3x + 3)}{h}$
now, clean it up and finish finding the limit which will give you the derivative of f.