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

Question

Ezra Adams · Accepted Answer

An equation of a tangent line is a linear approximation.

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]}

Isaiah Allen · Answer

undefined To find the linear approximation of the function $ f(x) = \frac{2}{x} $ at $ x_0 = 1 $, follow these steps:

1. Calculate the derivative of the function $ f(x) $.
2. Evaluate the derivative at the point $ x = x_0 $.
3. Use the derivative evaluated at $ x_0 $ to find the equation of the tangent line to the curve at $ x = x_0 $.
4. The equation of the tangent line represents the linear approximation of the function near the point $ x_0 $.

Elijah Abram · Answer

undefined To find the linear approximation of $ f(x) = \frac{2}{x} $ at $ x_0 = 1 $, you first find the derivative of $ f(x) $ with respect to $ x $, denoted as $ f'(x) $. Then, you evaluate $ f(x_0) $ and $ f'(x_0) $. Finally, you use the formula for linear approximation:

$$ L(x) = f(x_0) + f'(x_0)(x - x_0) $$

Substitute the values of $ f(x_0) $ and $ f'(x_0) $ into the formula, and simplify to find the linear approximation function $ L(x) $.