Determine The First Four Terms Of The Taylor Series For F ( X ) = E X F(x)=e^x F ( X ) = E X Centered At A = − 1 A=-1 A = − 1 . What Is The Coefficient Of ( X + 1 ) 3 (x+1)^3 ( X + 1 ) 3 In That Series?A. 1 6 E \frac{1}{6e} 6 E 1 ​ B. 1 6 \frac{1}{6} 6 1 ​ C. 6 E \frac{6}{e} E 6 ​ D.

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Determine the First Four Terms of the Taylor Series for f(x)=exf(x)=e^x Centered at a=−1a=-1

The Taylor series is a powerful tool in calculus for approximating functions. It is a way to represent a function as an infinite sum of terms, each of which is a polynomial. In this article, we will determine the first four terms of the Taylor series for the function f(x)=exf(x)=e^x centered at a=−1a=-1. We will also find the coefficient of (x+1)3(x+1)^3 in that series.

The Taylor series formula is given by:

f(x)=f(a)+f′(a)1!(x−a)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+⋯f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots

where f(x)f(x) is the function we want to approximate, aa is the center of the series, and f′(x)f'(x), f′′(x)f''(x), and f′′′(x)f'''(x) are the first, second, and third derivatives of f(x)f(x), respectively.

Finding the First Four Terms of the Taylor Series

To find the first four terms of the Taylor series for f(x)=exf(x)=e^x centered at a=−1a=-1, we need to find the values of f(−1)f(-1), f′(−1)f'(-1), f′′(−1)f''(-1), and f′′′(−1)f'''(-1).

First, we find the value of f(−1)f(-1):

f(−1)=e−1=1ef(-1) = e^{-1} = \frac{1}{e}

Next, we find the value of f′(−1)f'(-1):

f′(x)=exf'(x) = e^x

f′(−1)=e−1=1ef'(-1) = e^{-1} = \frac{1}{e}

Now, we find the value of f′′(−1)f''(-1):

f′′(x)=exf''(x) = e^x

f′′(−1)=e−1=1ef''(-1) = e^{-1} = \frac{1}{e}

Finally, we find the value of f′′′(−1)f'''(-1):

f′′′(x)=exf'''(x) = e^x

f′′′(−1)=e−1=1ef'''(-1) = e^{-1} = \frac{1}{e}

The First Four Terms of the Taylor Series

Now that we have found the values of f(−1)f(-1), f′(−1)f'(-1), f′′(−1)f''(-1), and f′′′(−1)f'''(-1), we can plug them into the Taylor series formula to find the first four terms of the series:

f(x)=1e+1e1!(x+1)+1e2!(x+1)2+1e3!(x+1)3+⋯f(x) = \frac{1}{e} + \frac{\frac{1}{e}}{1!}(x+1) + \frac{\frac{1}{e}}{2!}(x+1)^2 + \frac{\frac{1}{e}}{3!}(x+1)^3 + \cdots

Simplifying the expression, we get:

f(x)=1e+x+1e+(x+1)22e+(x+1)36e+⋯f(x) = \frac{1}{e} + \frac{x+1}{e} + \frac{(x+1)^2}{2e} + \frac{(x+1)^3}{6e} + \cdots

Finding the Coefficient of (x+1)3(x+1)^3

To find the coefficient of (x+1)3(x+1)^3, we need to look at the term in the Taylor series that contains (x+1)3(x+1)^3. In this case, it is the fourth term:

(x+1)36e\frac{(x+1)^3}{6e}

The coefficient of (x+1)3(x+1)^3 is therefore 16e\frac{1}{6e}.

In this article, we determined the first four terms of the Taylor series for f(x)=exf(x)=e^x centered at a=−1a=-1. We also found the coefficient of (x+1)3(x+1)^3 in that series, which is 16e\frac{1}{6e}. This result can be used to approximate the function f(x)=exf(x)=e^x near the point x=−1x=-1.

In our previous article, we determined the first four terms of the Taylor series for the function f(x)=exf(x)=e^x centered at a=−1a=-1. We also found the coefficient of (x+1)3(x+1)^3 in that series. In this article, we will answer some frequently asked questions about the Taylor series for f(x)=exf(x)=e^x centered at a=−1a=-1.

Q: What is the Taylor series formula?

A: The Taylor series formula is given by:

f(x)=f(a)+f′(a)1!(x−a)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+⋯f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots

Q: How do I find the first four terms of the Taylor series for f(x)=exf(x)=e^x centered at a=−1a=-1?

A: To find the first four terms of the Taylor series for f(x)=exf(x)=e^x centered at a=−1a=-1, you need to find the values of f(−1)f(-1), f′(−1)f'(-1), f′′(−1)f''(-1), and f′′′(−1)f'''(-1). You can then plug these values into the Taylor series formula to find the first four terms of the series.

Q: What is the coefficient of (x+1)3(x+1)^3 in the Taylor series for f(x)=exf(x)=e^x centered at a=−1a=-1?

A: The coefficient of (x+1)3(x+1)^3 in the Taylor series for f(x)=exf(x)=e^x centered at a=−1a=-1 is 16e\frac{1}{6e}.

Q: Can I use the Taylor series for f(x)=exf(x)=e^x centered at a=−1a=-1 to approximate the function f(x)=exf(x)=e^x near the point x=−1x=-1?

A: Yes, you can use the Taylor series for f(x)=exf(x)=e^x centered at a=−1a=-1 to approximate the function f(x)=exf(x)=e^x near the point x=−1x=-1. The Taylor series provides a good approximation of the function near the center of the series.

Q: How many terms of the Taylor series do I need to include to get a good approximation of the function f(x)=exf(x)=e^x near the point x=−1x=-1?

A: The number of terms of the Taylor series that you need to include to get a good approximation of the function f(x)=exf(x)=e^x near the point x=−1x=-1 depends on the desired level of accuracy. In general, you need to include more terms to get a more accurate approximation.

Q: Can I use the Taylor series for f(x)=exf(x)=e^x centered at a=−1a=-1 to find the value of the function f(x)=exf(x)=e^x at any point xx?

A: No, you cannot use the Taylor series for f(x)=exf(x)=e^x centered at a=−1a=-1 to find the value of the function f(x)=exf(x)=e^x at any point xx. The Taylor series is only valid near the center of the series, which is x=−1x=-1 in this case.

In this article, we answered some frequently asked questions about the Taylor series for f(x)=exf(x)=e^x centered at a=−1a=-1. We hope that this article has been helpful in understanding the Taylor series and its applications.

The final answer is 16e\boxed{\frac{1}{6e}}.