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

`f(x) = x^3`.

At `x = 2`, we have `y = 8`.

`f'(x) = 3x^2`, so at `x=2`, we have `f'(2) = 12`

The linearization is

`L(x) = 8+12(x-2)`

The equation of the tangent line

The line tangent to the graph at the point where `x=2` contains tha point `(2,8) and has slope`m=12#.

In point-slope form the equation is

`(y-8) = 12(x-2)`

When thinking of it as a linear approximation, we write

`y=8+12(x-2)`

More explanation

The linearization uses `y=8` as a starting point and adds the change in `y` along the tangent line for a particular change in `x`.

The change in `y` along the tangent line is called the differential of `y` and is denoted `dy`.
On a line of slope `m`, the change in `y` along the line is the slope times the change in `x`, in other words,
the change in `y` along the line `= m Deltax`

For the differential, we change the notation to `dx` and write:

`dy=mdx` where `m = f(x)` at some chosen `x=a`.

`dy = f'(a)Deltax = f'(a)(x-a)`

In this question the chosen value of `x` was `a = 2`,
so `dy = f'(a)(x-a) = 12(x-2)`.

The approximation begins at `y = 8` and adds `dy`

`L(x) = f(a) + f'(a)(x-a) = 8 + 12(x-2)`