How do you find the linearization of $f(x) = sqrt(x² + 2)$ at a=3?

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Answer 1 · 2016-08-31T12:44:24

The linearization (or linear approximation) of $f$ at $a$ is the equation of the tangent line at $x=a$ .

$f(x) = sqrt(x^2+2)$ at $a = 3$

$f(3) = sqrt11$ and

$f'(x) = x/sqrt(x^2+2)$ , so $m = f'(3) = 3/sqrt11$ .

The tangent line has point slope form

$y-sqrt11 = 3/sqrt11 (x-3)$ .

The linearization can be written in many ways, but one is

$L(x) = f(a)+f'(a)(x-a)$ .

In this case:

$L(x) = sqrt11 + 3/sqrt11 (x-a)$ .

How do you find the linearization of f(x) = sqrt(x² + 2)f(x) = sqrt(x² + 2) at a=3?