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

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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#