Question

How do you find the linearization at x=1 of `f(x)= 2x^3-3x`?

Answer 1

The linearization is given by `3x-4`.

Explanation:

The linearization of a function `f` at a certain point `x_0` is the tangent line to `f` in `x_0`

It is given by `f(x_0)+f'(x_0)(x-x_0)`.

In your case, `f'(x)=6x^2-3`, and thus `f'(1)=6-3=3`

Your line is thus

`f(1)+f'(1)(x-1)=-1+3(x-1)=-1+3x-3=3x-4`.

Here you can see the graphs of the function and the line, checking that it is indeed the tangent in `x=1`.