# What is the Taylor Series generated by #f(x) = x - x^3#, centered around a = -2?

The answer is

There are two methods to get the answer.

Here are the details, which follow from the binomial theorem (Pascal's triangle ).

The answer is therefore, as we saw before,

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The Taylor series generated by ( f(x) = x - x^3 ) centered around ( a = -2 ) is:

[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots ]

First, find the derivatives of ( f(x) ) up to the third derivative:

( f(x) = x - x^3 )

( f'(x) = 1 - 3x^2 )

( f''(x) = -6x )

( f'''(x) = -6 )

Now, evaluate these derivatives at ( a = -2 ):

( f(-2) = -2 - (-2)^3 = -2 + 8 = 6 )

( f'(-2) = 1 - 3(-2)^2 = 1 - 12 = -11 )

( f''(-2) = -6(-2) = 12 )

( f'''(-2) = -6 )

Now, plug these values into the Taylor series formula:

[ f(x) = 6 - 11(x+2) + \frac{12}{2!}(x+2)^2 - \frac{6}{3!}(x+2)^3 + \cdots ]

This is the Taylor series representation of ( f(x) = x - x^3 ) centered around ( a = -2 ).

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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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