Find the slope of f(x) = x 3−3x at the point x = 2 and the corresponding equation of the tangent line. Use the definition of the slope of graph at a point.
I got to lim h=0 (x+h)^3-3(x+h)-(x^3-3x)/h)
+600 $\lim_{h\to 0}\frac{(x+h)^3-3(x+h)-x^3+3x}{h}=\lim_{h\to 0}\frac{-3 h + h^3 + 3 h^2 x + 3 h x^2}{h}=\lim_{h\to 0} -3+h^2+3hx+3x^2=3x^2-3$
At $x=2$, this is $3*4-3=9$.
Using this, you can easily find the tangent line equation so I will leave that to you.
+600 $\lim_{h\to 0}\frac{(x+h)^3-3(x+h)-x^3+3x}{h}=\lim_{h\to 0}\frac{-3 h + h^3 + 3 h^2 x + 3 h x^2}{h}=\lim_{h\to 0} -3+h^2+3hx+3x^2=3x^2-3$
At $x=2$, this is $3*4-3=9$.
Using this, you can easily find the tangent line equation so I will leave that to you.
