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

Question

1994 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-25T15:46:25

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

$f(x) =sqrt(x² + 2)$

$f(3) = sqrt11$

$f'(x) = x/sqrt(x^2+2)$ , so

$f'(3) = 3/sqrt11$

The linearization/tangent line can be written:

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

Or as $y = sqrt11 + 3/sqrt11(x-3)$

Or, in slope intercept for as

$y = 3/sqrt11x-9/sqrt11+sqrt11$

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