Find the slopes of the tangent line to the curve \(\displaystyle y=x^3-3x\) at the points where x=-2,-1,0,1,2.

So \(\displaystyle m_{tan}=\frac{[(c+h)^3-3(c+h)]-(c^3-3c)}{h}\)

\(\displaystyle =\frac{(c^3+3c^2h+3ch^2+h^3)-3c-3h-c^3+3c)}{h}\)

\(\displaystyle =\frac{3c^2h+3ch^2-3h+h^3}{h}\)

\(\displaystyle =\frac{h(3c^2+3ch-3+h^2)}{h}\)

lim h->0 \(\displaystyle 3c^2+3ch-3+h^2=3c^2-3\)

so at x=-2, \(\displaystyle m_{tan}=9\)

x=-1, \(\displaystyle m_{tan}=0\)

x=0, \(\displaystyle m_{tan}=-3\)

x=1, \(\displaystyle m_{tan}=0\)

x=2, \(\displaystyle m_{tan}=9\)

Thank you for any help!