Linear Approximation

Question

The linear approximation at  x = 0 to f(x) = 1/sqrt(3-x)  is L(x) = A + Bx

where

A = ?

B = ?

Answer 1

Generally speaking, the linear approximation, L(x) of a function f(x) at a point x = a is:

L(x) = f(a) + f '(a)(x - a)

In this case, where a = 0,then the pieces would be:

f(0) = 1/√(3 - 0) = 1/√3 = √3/3

To find f '(a), we need to find f '(x). So let's write f(x) as (3 - x)^-1/2 then using the power rule and the chain rule we get f '(x) = (-)(-1/2)(3 - x)^-3/2 and plugging in 0, we get f '(0) = 1/2(3)^-3/2 = 1/(2√27) = 1/(6√3) = √3/18

So, plugging this in, we get:

L(x) = √3/3 + √3/18x

So A = √3/3 and B = √3/18

Answer 2

y = (3-x)^(-1/2)

y' = (-1/2) (3-x)^(-3/2) (-1)

= (1/2)(3-x)^(-3/2)

at x=0, the slope of the tangent line is

f'(0) = (1/2)(3)^(-3/2) = (1/2)(1 / sqrt(3)^3) = 1 / [2 * 3 ^(3/2) ] = k

f(0) = 1/sqrt(3) =

B = y-mx = 1/ sqrt(3) - (1/2) ( 1 / sqrt(3)^3)(0) = 1/ sqrt(3

y = mx + b = 1 / [2 * 3 ^(3/2) ] x + 1/ sqrt(3)

= sqrt(3)/3 [ (1/54) x + 1 ]