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

1 Answer
Jim H
Apr 21, 2016

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]}

Related questions
See all questions in Using the Tangent Line to Approximate Function Values
Impact of this question
4377 views around the world
You can reuse this answer
Creative Commons License