Answer:
An improper fraction
Step-by-step explanation:
In improper fraction is when the numerator (top number) is greater than or equal to the denominator (bottom number)
Taylor series is ![f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }](https://tex.z-dn.net/?f=f%28x%29%20%3D%20ln2%20%2B%20%5Csum_%7Bn%3D1%29%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn-1%7D%20%5Cfrac%7B%28n-1%29%21%7D%7Bn%21%289%29%5E%7Bn%7D%28x9%29%5E%7B2%7D%20%20%7D)
To find the Taylor series for f(x) = ln(x) centering at 9, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have
f(x) = ln(x)
![f^{1}(x)= \frac{1}{x} \\f^{2}(x)= -\frac{1}{x^{2} }\\f^{3}(x)= -\frac{2}{x^{3} }\\f^{4}(x)= \frac{-6}{x^{4} }](https://tex.z-dn.net/?f=f%5E%7B1%7D%28x%29%3D%20%5Cfrac%7B1%7D%7Bx%7D%20%5C%5Cf%5E%7B2%7D%28x%29%3D%20-%5Cfrac%7B1%7D%7Bx%5E%7B2%7D%20%7D%5C%5Cf%5E%7B3%7D%28x%29%3D%20-%5Cfrac%7B2%7D%7Bx%5E%7B3%7D%20%7D%5C%5Cf%5E%7B4%7D%28x%29%3D%20%5Cfrac%7B-6%7D%7Bx%5E%7B4%7D%20%7D)
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Since we need to have it centered at 9, we must take the value of f(9), and so on.
f(9) = ln(9)
![f^{1}(9)= \frac{1}{9} \\f^{2}(9)= -\frac{1}{9^{2} }\\f^{3}(x)= -\frac{1(2)}{9^{3} }\\f^{4}(x)= \frac{-1(2)(3)}{9^{4} }](https://tex.z-dn.net/?f=f%5E%7B1%7D%289%29%3D%20%5Cfrac%7B1%7D%7B9%7D%20%5C%5Cf%5E%7B2%7D%289%29%3D%20-%5Cfrac%7B1%7D%7B9%5E%7B2%7D%20%7D%5C%5Cf%5E%7B3%7D%28x%29%3D%20-%5Cfrac%7B1%282%29%7D%7B9%5E%7B3%7D%20%7D%5C%5Cf%5E%7B4%7D%28x%29%3D%20%5Cfrac%7B-1%282%29%283%29%7D%7B9%5E%7B4%7D%20%7D)
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Following the pattern, we can see that for
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This applies for n ≥ 1, Expressing f(x) in summation, we have
![\sum_{n=0}^{\infinite} \frac{f^{n}(9) }{n!} (x-9)^{2}](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfinite%7D%20%5Cfrac%7Bf%5E%7Bn%7D%289%29%20%7D%7Bn%21%7D%20%28x-9%29%5E%7B2%7D)
Combining ln2 with the rest of series, we have
![f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }](https://tex.z-dn.net/?f=f%28x%29%20%3D%20ln2%20%2B%20%5Csum_%7Bn%3D1%29%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn-1%7D%20%5Cfrac%7B%28n-1%29%21%7D%7Bn%21%289%29%5E%7Bn%7D%28x9%29%5E%7B2%7D%20%20%7D)
Taylor series is ![f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }](https://tex.z-dn.net/?f=f%28x%29%20%3D%20ln2%20%2B%20%5Csum_%7Bn%3D1%29%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn-1%7D%20%5Cfrac%7B%28n-1%29%21%7D%7Bn%21%289%29%5E%7Bn%7D%28x9%29%5E%7B2%7D%20%20%7D)
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Answer:
D
21/7 is basically 1/3 so it would be C
Step-by-step explanation:
10. Fish don't drown (; nice try tho