Question

How do you find the linearization at x=0 of `f ' (x) = cos (x^2)`?

Answer 1

Use the fact that the linearization at `a` is the tangent line: `L(x) = f(a)+f'(a)(x-a)`

Explanation:

we want to linearize `f(x) = cos(x^2)` at `x=0`

`f(0) = cos(0^2) = 1`

`f'(x) = -2xsin(2x)`, so `f'(0) = 0`.

The linearization is `L(x) = 1+0(x-0) = 1`.

The linearization is the horizontal line `y=1`.

To help clarify the situation, here's the graph of `y=cos(x^2)`.

Observe that near `0` the graph is nearly the horisontal line `y=1`.

(You can zoom in/out and drag the image around. When you leave the page and return the default view will be back.)

graph{y=cos(x^2) [-4.15, 4.62, -1.61, 2.775]}

Here is a graph with both `y=cos(x^2)` and `y=1`
The more you zoom in, the closer the two graphs look. When you zoom in enough, you can't even see that there are two graphs. The same thing happens at `x=sqrt(2pi) ~~ 2.506` or and maximum or minimum, but you have to zoom in a lot more.)

graph{(y-cos(x^2))(y-1)=0 [-4.15, 4.62, -1.61, 2.775]}