Question

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

Answer 1

First, you have to differentiate `f(x)` using the chain rule.

Explanation:

Let `y = u^3` and `u = x - 1`.

`y' = 3u^2` and `u' = 1`

`dy/dx = 3u^2 xx 1`

`dy/dx = 3(x - 1)^2`

Next, we have to find the point through which our tangent, and our function pass.

`f(2) = (2 - 1)^3`

`f(2) = 1`

So, the function and the tangent pass through the point `(2, 1)`.

Now, we need to find the slope of the tangent by plugging in and evaluating `x = a` inside the derivative.

`m_"tangent" = 3(2 - 1)^2`

`m_"tangent" = 3`

We must next determine the equation of the tangent using point-slope form.

`y - y_1 = m(x - x_1)`

`y - 1 = 3(x - 2)`

`y - 1 = 3x - 6`

`y = 3x - 5`

Hence, the equation of the tangent line to `f(x) = (x - 1)^3` at the point `x= 2` is `y = 3x - 5`.

Hopefully this helps!