Question #1d4dc

Question

1850 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-27T02:41:33

There is more than one method called the three-step method. (The steps are described differently.) They have the following in common.

$f(x) = 2x^3$

We'll need to find $f(x+h)$ and $f(x+h)-f(x)$ and $(f(x+h)-f(x))/h$

Some call this steps 1 and 2, others call all of this step 1.

$f(x+h) = 2(x+h)^3 = 2(x^3+3x^2h+3xh^2+h^3)$

$= 2x^3+6x^2h+6xh^2+2h^3$

$f(x+h)-f(x) = 2(x+h)^3 - 2x^3$

$= 2x^3+6x^2h+6xh^2+2h^3-2x^3$

$= 6x^2h+6xh^2+2h^3$

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

$= (h(6x^2+6xh+2h^2))/h$

$= 6x^2+6xh+2h^2$

Now we need to find $lim_(hrarr0)(f(x+h)-f(x))/h$

$lim_(hrarr0)(6x^2+6xh+2h^2) = 6x^2+0+0 = 6x^2$

So $f'(x) = 6x^2$