How do you find the linearization of #f(x) = 2x³ + 4x² + 6# at a=3?

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Answer 1 · 2017-03-07T14:33:42

#f(x) approx 96 + 78 (x-3) #

By linearising you are saying that, around #x = a#:

#f(x) approx f(a) + (x-a) f'(a)#

Or:

#L(x) = f(a) + (x-a) f'(a)#

Here:

#f(a) = 96#

#f'(x) = 6x^2 + 8 x#

#f'(a) = 78#

So that means:

#L(x) = 96 + 78 (x-3) #

Try it by plugging some values around #x = 3# into both #f(x)# and #L(x)# to see if it works.