Question 1038078
Here is the second one.
y=x^3-bx
f(x+h)-f(x)/h
numerator=(x+h)^3-x^3=x^3+3x^2h+3xh^2+h^3-b(x+h)-x^3-bx
This is 3x^2h+3xh^2+h^3-bx-bh-bx=3x^2h+3xh^2+h^3-bh
Divide that by h
=3x^2+3xh+h^2-b
as h goes to 0, f'(x^3) goes to 3x^2-b, which is the derivative.
f(x) when x=2 is 8-2b
The derivative must equal zero, so 3x^2-b=0 for x=2.  Therefore b=12
the y-coordinate of the point when x=2 is 8-24 or -16.  The point is (2,-16)
{{{graph(300,200,-10,10,-30,30,x^3-12x,-16)}}}