Linearizations of a
function are
lines—usually lines that can be used for purposes of calculation. Linearization is an effective method for approximating the output of a function y = f(x) at any x = a based on the value and
slope of the function at x = b, given that f(x) is differentiable on [a, b] (or [b, a]) and that a is close to b. In short, linearization approximates the output of a function near x = a. For example, \sqrt{4} = 2. However, what would be a good approximation of \sqrt{4.001} = \sqrt{4 + .001}? For any given function y = f(x), f(x) can be approximated if it is near a known differentiable point. The most basic requisite is that L_a(a) = f(a), where L_a(x) is the linearization of f(x) at x = a. The
point-slope form of an equation forms an equation of a line, given a point (H, K) and slope M. The general form of this equation is: y - K = M(x - H). Using the point (a, f(a)), L_a(x) becomes y = f(a) + M(x - a). Because differentiable functions are
locally linear, the best slope to substitute in would be the slope of the line
tangent to f(x) at x = a. While the concept of local linearity applies the most to points
arbitrarily close to x = a, those relatively close work relatively well for linear approximations. The slope M should be, most accurately, the slope of the tangent line at x = a. Visually, the accompanying diagram shows the tangent line of f(x) at x. At f(x+h), where h is any small positive or negative value, f(x+h) is very nearly the value of the tangent line at the point (x+h, L(x+h)). The final equation for the linearization of a function at x = a is: y = (f(a) + f'(a)(x - a)) For x = a, f(a) = f(x). The
derivative of f(x) is f'(x), and the slope of f(x) at a is f'(a).
Example To find \sqrt{4.001}, we can use the fact that \sqrt{4} = 2. The linearization of f(x) = \sqrt{x} at x = a is y = \sqrt{a} + \frac{1}{2 \sqrt{a}}(x - a), because the function f'(x) = \frac{1}{2 \sqrt{x}} defines the slope of the function f(x) = \sqrt{x} at x. Substituting in a = 4, the linearization at 4 is y = 2 + \frac{x-4}{4}. In this case x = 4.001, so \sqrt{4.001} is approximately 2 + \frac{4.001-4}{4} = 2.00025. The true value is close to 2.00024998, so the linearization approximation has a relative error of less than 1 millionth of a percent. ==Linearization of a multivariable function==