2.7. Linearization and Newton’s Method http://www.ck12.org
is a linearization offnearx 0.
Example 1:
Find the linearization off(x) =√x+3 at pointx=1.
Solution:
Taking the derivative off(x),
f′(x) =^12 (x+ 3 )−^1 /^2 ,we havef( 1 ) =
√
4 = 2 ,f′( 1 ) = 1 / 4 ,andf(x)≈f(x 0 )+f′(x 0 )(x−x 0 )
≈ 2 +^14 (x− 1 )
≈^14 x+^74.This tells us that near the pointx=1, the functionf(x) =
√
x+3 approximates the liney= (x/ 4 ) + 7 /4. As we
move away fromx=1, we lose accuracy (Figure 9).
Figure 9
Example 2:
Find the linearization ofy=sinxatx=π/3.
Solution:
Sincef(π/ 3 ) =sin(π/ 3 ) =
√
3 /2, andf′(x) =cosx,f′(π/ 3 ) =cos(π/ 3 ) = 1 / 2 ,we have