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?

Question

Ezekiel Alexander · Accepted Answer

#-12.#

#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.#

Christopher Agresta · Answer

undefined To find the slope of the tangent line to the graph y=1-x^3 at x=2 using the limit definition, we can follow these steps:

1. Start with the equation of the graph: y = 1 - x^3.

2. Determine the derivative of the function y with respect to x, denoted as dy/dx or f'(x). In this case, the derivative is f'(x) = -3x^2.

3. Substitute the given x-value, x=2, into the derivative function to find the slope at that point: f'(2) = -3(2)^2 = -12.

4. The slope of the tangent line to the graph y=1-x^3 at x=2 is -12.