How do you find the linear approximation f(x)=2/xf(x)=2x, x_0=1x0=1?

1 Answer
Jim H
May 18, 2017

An equation of a tangent line is a linear approximation.

Explanation:

The at a chosen value of x = x_0x=x0, the function has value f(x_0)f(x0)

The tangent line through (x_0,f(x_0))(x0,f(x0)) 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]}

Related questions
See all questions in Using the Tangent Line to Approximate Function Values
Impact of this question
5763 views around the world
You can reuse this answer
Creative Commons License