Editing Linearization (section)

==Linearization of a function==
Linearizations of a [[function (mathematics)|function]] are [[linear function|lines]]—usually lines that can be used for purposes of calculation. Linearization is an effective method for approximating the output of a function <math>y = f(x)</math> at any <math>x = a</math> based on the value and [[slope]] of the function at <math>x = b</math>, given that <math>f(x)</math> is differentiable on <math>[a, b]</math> (or <math>[b, a]</math>) and that <math>a</math> is close to <math>b</math>. In short, linearization approximates the output of a function near <math>x = a</math>.

For example, <math>\sqrt{4} = 2</math>. However, what would be a good approximation of <math>\sqrt{4.001} = \sqrt{4 + .001}</math>?

For any given function <math>y = f(x)</math>, <math>f(x)</math> can be approximated if it is near a known differentiable point. The most basic requisite is that <math>L_a(a) = f(a)</math>, where <math>L_a(x)</math> is the linearization of <math>f(x)</math> at <math>x = a</math>. The [[Linear equation#Point–slope form|point-slope form]] of an equation forms an equation of a line, given a point <math>(H, K)</math> and slope <math>M</math>. The general form of this equation is: <math>y - K = M(x - H)</math>.

Using the point <math>(a, f(a))</math>, <math>L_a(x)</math> becomes <math>y = f(a) + M(x - a)</math>. Because differentiable functions are [[Local linearity|locally linear]], the best slope to substitute in would be the slope of the line [[tangent]] to <math>f(x)</math> at <math>x = a</math>.

While the concept of local linearity applies the most to points [[Limit of a function#Limit of a function at a point|arbitrarily close]] to <math>x = a</math>, those relatively close work relatively well for linear approximations. The slope <math>M</math> should be, most accurately, the slope of the tangent line at <math>x = a</math>.

[[Image:Tangent-calculus.svg|thumb|300px|An approximation of ''f''(''x'') = ''x''<sup>2</sup> at (''x'', ''f''(''x''))]]
Visually, the accompanying diagram shows the tangent line of <math>f(x)</math> at <math>x</math>. At <math>f(x+h)</math>, where <math>h</math> is any small positive or negative value, <math>f(x+h)</math> is very nearly the value of the tangent line at the point <math>(x+h, L(x+h))</math>.

The final equation for the linearization of a function at <math>x = a</math> is:
<math display="block">y = (f(a) + f'(a)(x - a))</math>

For <math>x = a</math>, <math>f(a) = f(x)</math>. The [[derivative]] of <math>f(x)</math> is <math>f'(x)</math>, and the slope of <math>f(x)</math> at <math>a</math> is <math>f'(a)</math>.

===Example===
To find <math>\sqrt{4.001}</math>, we can use the fact that <math>\sqrt{4} = 2</math>. The linearization of  <math>f(x) = \sqrt{x}</math> at <math>x = a</math> is <math>y = \sqrt{a} + \frac{1}{2 \sqrt{a}}(x - a)</math>, because the function <math>f'(x) = \frac{1}{2 \sqrt{x}}</math> defines the slope of the function <math>f(x) = \sqrt{x}</math> at <math>x</math>. Substituting in <math>a = 4</math>, the linearization at 4 is <math>y = 2 + \frac{x-4}{4}</math>. In this case <math>x = 4.001</math>, so <math>\sqrt{4.001}</math> is approximately <math>2 + \frac{4.001-4}{4} = 2.00025</math>. The true value is close to 2.00024998, so the linearization approximation has a relative error of less than 1 millionth of a percent.