Question 1087244
<font color="black" face="times" size="3">The first thing we need to do is find {{{f(x+h)}}}. We do this by replacing every copy of x with x+h


{{{f(x)= 2x^3 - x}}}


{{{f(x)= 2(x)^3 - (x)}}}


{{{f(highlight(x))= 2(highlight(x))^3 - (highlight(x))}}} Highlighting the x terms we will replace


{{{f(highlight(x+h))= 2(highlight(x+h))^3 - (highlight(x+h))}}} Replacing those terms with x+h


---------------------------------------


So we have


{{{f(x+h)= 2(x+h)^3 - (x+h)}}}


Let's expand that out and simplify


{{{f(x+h)= 2(x+h)*(x+h)^2 - (x+h)}}}


{{{f(x+h)= (x+h)(2*(x+h)^2 - 1)}}}


{{{f(x+h)= (x+h)(2*(x^2+2xh+h^2) - 1)}}}


{{{f(x+h)= (x+h)(2x^2+4xh+2h^2 - 1)}}}


{{{f(x+h)= x(2x^2+4xh+2h^2 - 1)+h(2x^2+4xh+2h^2 - 1)}}}


{{{f(x+h)= 2x^3+4x^2h+2h^2x - x+2x^2h+4xh^2+2h^3 - h}}}


{{{f(x+h)= 2x^3+4x^2h+2x^2h+2h^2x+4xh^2 - x+2h^3 - h}}}


{{{f(x+h)= 2x^3+6x^2h+6xh^2 - x+2h^3 - h}}}


---------------------------------------


Now we can compute the difference quotient


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


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


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


{{{(f(x+h)-f(x))/h = (h(6x^2+6xh+2h^2 - 1))/h}}}


{{{(f(x+h)-f(x))/h = (highlight(h)(6x^2+6xh+2h^2 - 1))/(highlight(h))}}}


{{{(f(x+h)-f(x))/h = (cross(h)(6x^2+6xh+2h^2 - 1))/(cross(h))}}}


{{{(f(x+h)-f(x))/h = 6x^2+6xh+2h^2 - 1}}}


---------------------------------------
---------------------------------------


So the final answer is {{{(f(x+h)-f(x))/h = 6x^2+6xh+2h^2 - 1}}}


Side Note: After this step is reached, you can plug in h = 0 to find the derivative of f(x). This topic probably hasn't come up yet in your course, but it definitely will come up in calculus. </font>