How do you find the linearization at x=2 of $f (x) = 3 x - \frac{2}{x^{2}}$ $f(x) = 3x - 2/x^2$ ?

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Answer 1 · 2017-06-11T12:51:30

Taylor: $\frac{11}{2} + \frac{13}{4} (x - 2) + O {(x - 2)}^{2}$ $11/2 + 13/4(x - 2) + O(x-2)^2$

$f'(x) = 3 + 2/x^3 => f'(2) = 3 + 2/8 = 13/4$

$L(x) = 13/4x + b and L(2) = f(2)$

$f(2) = 6 - 2/4 = 11/2$

$13/4 * 2 + b = 11/2 => b = 1$

The tangent line at x = 2 is $L(x) = 13/4 x + 1$

How do you find the linearization at x=2 of f(x) = 3x - 2/x^2f(x)=3x−2x2f(x) = 3x - 2/x^2?