Question

How do you find the linear approximation `f(x)=2/x`, `x_0=1`?

Answer 1

An equation of a tangent line is a linear approximation.

Explanation:

The at a chosen value of `x = x_0`, the function has value `f(x_0)`

The tangent line through `(x_0,f(x_0))` has slope `f'(x_0)`

The linear equation can be thought of as an equation for the tangent line or as a linear function giving an approximation of `f`.

The point-slope form of the equation of the tangent line (the line through `(x_0,f(x_0))` with slope `f'(x_0)`) is:

`y-y_0 = f'(x) (x-x_0)`

The linear approximation for `f` at `x_0` is

`y= y_0 + f'(x) (x-x_0)`

For `f(x) = 2/x` at `x_0=1` we get

`f(1) = 2` and `f'(x) = -2/x^2` so `m = f'(1) = -2`.

The linear approximation is

`y = 2-2(x-1)`

which you may prefer to write as

`y = -2x+4`.

Here is a graph showing both. You can zoom in and out and drag the graph around the window. When you navigate away from this answer and return, the graph will start n the same position it has now.

graph{(y-2/x)(y+2x-4)=0 [-0.58, 2.458, 1.252, 2.771]}