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)#.