Question

What is the equation of the tangent line of `f(x)=(x-1)^3` at `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.

f'(2) is the value of the tangent gradient and f(2) is the value of the y-coordinate , given x-coordinate.

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

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

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

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

hence 'partial' equation is y = 3x + c

now `f(2) = (3-2)^2 = 1 → (2,1)" is point on tangent "`

using (2,1) in y = 3x + c : 6 + c = 1 → c = -5

and equation of tangent is therefore : y = 3x - 5