Question

What is the local linearization of `y = (7+5x^2)^(-1/2)` at a=0?

Answer 1

It is `L(x)=7^(-1/2)` (or `L(x) = 1/sqrt7` if you prefer.)

Explanation:

We have `f(x) = (7+5x^2)^(-1/2)`, so

`f'(x) = (-1/2)(7+5x^2)^(-3/2)(10x)`.

At the chosen point, the derivative `f'(a)` is `0`.

So the liearization is horizontal: `L(x)=f(a) = 7^(-1/2)`

General Case
The local linearization (aka linear approximation), of function `f` at point `x=a` is a form of the equation for the tangent line at `(a,f(a))`.

We have `y=f(x)`, and we note that the tangent at the point where `x=a` intersects the graph at `(a,f(a))` and has slope `m=f'(a)`.

Writing the equation of the tangent line in point-slope form, we get

`y-f(a)=f'(a)(x-a)`.

Adding `f(a)` to both sides get us the local linearization (linear approximation) of `f(x)` at `a`:

`L(x)~~f(a)+f'(a)(x-a)`,

Here is a picture of the graph of the function in this question. As you zoom in centered at `x=0`, you'll see that the graph looks very much like a horizontal line.

graph{y=(7+5x^2)^(-1/2) [-3.467, 3.465, -1.35, 2.117]}