http://www.ck12.org Chapter 2. Derivatives
f(x)≈√ 3
2 +
1
2
(
x−π 3)
≈
√ 3
2 +
x
2 −π
6
≈x 2 + 0. 343.Figure 10
Newton’s Method
When faced with a mathematical problem that cannot be solved with simple algebraic means, such as finding the
roots of the polynomialx^3 − 2 x+ 3 = 0 ,calculus sometimes provides a way of finding the approximate solutions.
Let’s say we are interested in computing
√
5 without using a calculator or a table. To do so, think about this problem
in a different way. Assume that we are interested in solving the quadratic equation
f(x) =x^2 − 5 = 0which leads to the rootsx=±√5.
The idea here is to find the linearization of the above function, which is a straight-line equation, and then solve the
linear equation forx.
Since
√
4 <√
5 <
√
9
or
2 <
√
5 < 3 ,
We choose the linear approximation off(x)to be nearx 0 =2. Sincef(x) =x^2 − 5 ,f′(x) = 2 xand thusf( 2 ) =− 1
andf′( 2 ) = 4 .Using the linear approximation formula,