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Find the first five non-zero terms of the Taylor series for f(x)= -8eˣ centered at x=7

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Answer:

Explanation:

The Taylor series for a function f(x) centered at x=a is given by:

f(x) = f(a) + f'(a)(x-a)/1! + f''(a)(x-a)²/2! + f'''(a)(x-a)³/3! + ...

In this case, we need to find the Taylor series for f(x) = -8eˣ centered at x=7.

To find the first five non-zero terms, we need to evaluate the function and its derivatives at x=7.

First, let's find the derivatives of f(x) = -8eˣ: f'(x) = -8eˣ f''(x) = -8eˣ f'''(x) = -8eˣ f''''(x) = -8eˣ

Next, let's evaluate these derivatives at x=7: f(7) = -8e⁷ f'(7) = -8e⁷ f''(7) = -8e⁷ f'''(7) = -8e⁷ f''''(7) = -8e⁷

Now, we can plug these values into the Taylor series formula: f(x) = f(7) + f'(7)(x-7)/1! + f''(7)(x-7)²/2! + f'''(7)(x-7)³/3! + f''''(7)(x-7)⁴/4! + ...

Substituting the values we found: f(x) = -8e⁷ - 8e⁷(x-7)/1! - 8e⁷(x-7)²/2! - 8e⁷(x-7)³/3! - 8e⁷(x-7)⁴/4! + ...

To find the first five non-zero terms, we expand the series up to the fourth power term: f(x) = -8e⁷ - 8e⁷(x-7)/1! - 8e⁷(x-7)²/2! - 8e⁷(x-7)³/3!

Therefore, the first five non-zero terms of the Taylor series for f(x) = -8eˣ centered at x=7 are: -8e⁷ -8e⁷(x-7) -4e⁷(x-7)² -8/3 e⁷(x-7)³

I hope this helps! Let me know if you have any further questions.

User Pat Zabawa
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