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

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Answer 1 · 2015-11-09T02:12:37

$y = (6a^2+8a)x - (4a^3+4a^2-6)$

$f'(a)=6a^2+8a$

The tangent line has slope $m=f'(a)$ and goes through the point $(a,f(a))$ . So the tangent line has point slope form:

$y-f(a) = f'(a)(x-a)$ .

The linearization at $x=a$ , is:

$y = f'(a)x + f(a)-af'(a)$

$= (6a^2+8a)x - (4a^3+4a^2-6)$

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