Question

How do you use the limit definition to find the slope of the tangent line to the graph `y=1-x^3` at x=2?

Answer 1

`-12.`

Explanation:

`f(x)=1-x^3`

`f'(x)=lim_(h->0)(f(x+h)-f(x))/h`

`f(x+h)=1-(x+h)^3=1-(x^3+3x^2h+3xh^2+h^3)=1-x^3-3x^2h-3xh^2-h^3`

So,

`f'(x)=lim_(h->0)(1-x^3-3x^2h-3xh^2-h^3-(1-x^3))/h=lim_(h->0)(-3x^2h-3xh^2-h^3)/h=lim_(h->0)(h(-3x^2-3xh-h^2))/h=lim_(h->0)-3x^2-3xh-h^2=-3x^2`

So, `f'(x)=-3x^2.`

This gives us the rate of change, IE slope of the tangent line to the curve, of `f(x)=1-x^3` at any point `x.` If we want the slope of the tangent line at `x=2,` we calculate `f'(2)=-3(2^2)=-3(4)=-12.`