Question

How do you find the equation of the tangent line to the curve `f(x) = (x-1)^3` at the point where x = 2?

Answer 1

y = 3x - 5

Explanation:

To find the equation in the form y = mx + c , where m represents the gradient and c, the y-intercept.
The value of f'(2) will give m and f(2) will assist in finding c.

differentiate using the `color(blue)" chain rule "`

`d/dx f(g(x)) = f'(g(x)) . g'(x)`

hence f'(x) `=3(x -1)^2 d/dx(x - 1 ) = 3(x - 1 )^2`

f'(2) = `3(1)^2 = 3 " gradient of tangent "`

and f(2) =`(1)^3 = 1 rArr (2,1) " is tangent point "`

partial equation of tangent is y = 3x + c , and using (2,1 )

gives : 1 =3(2) + c → c = - 5

thus equation of tangent is : y = 3x - 5