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

1 Answer
Jan 16, 2016

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]}