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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.
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