Question

How do you find the linearization of `f(x)=4x^3-5x-1` at a=2?

Answer 1

Find the equation of the tangent line at the point `(2,f(2))`

Explanation:

`f(2) = 21`

`f'(x) = 12x^2-5`, so `f'(2) = 43`

The equation of the tangent line (in point-slope form) is

`y-21=43(x-2)`.

The linearization is

`L(x) = 21 + 43(x-2)`.