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

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To find the linearization of ( f(x) = 2x^3 + 4x^2 + 6 ) at ( a = 3 ), we first find the derivative of ( f(x) ) which is ( f'(x) = 6x^2 + 8x ).

Then, we evaluate ( f(a) ) and ( f'(a) ) at ( a = 3 ):

( f(3) = 2(3)^3 + 4(3)^2 + 6 = 2(27) + 4(9) + 6 = 54 + 36 + 6 = 96 )

( f'(3) = 6(3)^2 + 8(3) = 6(9) + 24 = 54 + 24 = 78 )

The linearization of ( f(x) ) at ( a = 3 ) is given by ( L(x) = f(a) + f'(a)(x - a) ), so substituting the values we found:

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

Therefore, the linearization of ( f(x) = 2x^3 + 4x^2 + 6 ) at ( a = 3 ) is ( L(x) = 78x - 186 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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