Question

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

Answer 1

First of all, let's evaluate the function in `x=2`: we have `(2-1)^3=1^3=1`.

So, we are looking for a line passing through the point `(2,1)`. The general equation of a line passing through a point `(x_0,y_0)` is
`y-y_0=m(x-x_0)`. So, in our, case, we have the equation
`y-1=m(x-2)`.

Now we need to find the slope `m`, which is by definition the value of the derivative in `x=2`.

Using the power rule, which states that `D f^n(x)= n f^{n-1}(x) f'(x)`, we have that `f'(x)=3(x-1)^2`. This means that `f'(2)=3`.

We thus have everything we need to answer: the line is
`y-1=3(x-2)`, which we can bring into the standard form writing
`y=3x-5`

Here's a link where you can see the graph.