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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.
==Linearization of a multivariable function{{anchor|Multivariable functions}}==
{{See also|Taylor series#In several variables}}
The equation for the linearization of a function <math>f(x,y)</math> at a point <math>p(a,b)</math> is:
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:<math>f({\mathbf{x}}) \approx f({\mathbf{p}}) + \left. {\nabla f} \right|_{\mathbf{p}} \cdot ({\mathbf{x}} - {\mathbf{p}})</math>
where <math>\mathbf{x}</math> is the vector of variables, <math>{\nabla f}</math> is the [[gradient]], and <math>\mathbf{p}</math> is the linearization point of interest
.<ref>[http://www.ece.jhu.edu/~pi/Courses/454/linear.pdf Linearization. The Johns Hopkins University. Department of Electrical and Computer Engineering] {{webarchive|url=https://web.archive.org/web/20100607120539/http://www.ece.jhu.edu/~pi/Courses/454/linear.pdf |date=2010-06-07 }}</ref>
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