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.

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