1. let ( f ) be a function with third - derivative ( f(x)=2x^{-3}). what is the coefficient of ( (x - 1)^4 )…

1. let ( f ) be a function with third - derivative ( f(x)=2x^{-3}). what is the coefficient of ( (x - 1)^4 ) in the fourth - degree taylor polynomial of ( f ) about ( x = 1)? (a) (-\frac{1}{4}) (b) (-6) (c) (-\frac{3}{2}) (d) (\frac{3}{2})

1. let ( f ) be a function with third - derivative ( f(x)=2x^{-3}). what is the coefficient of ( (x - 1)^4 ) in the fourth - degree taylor polynomial of ( f ) about ( x = 1)? (a) (-\frac{1}{4}) (b) (-6) (c) (-\frac{3}{2}) (d) (\frac{3}{2})

Answer

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of a function (f(x)) about (x = a) is given by (f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x - a)^{n}), where (f^{(n)}(a)) is the (n) - th derivative of (f) evaluated at (x=a). For the fourth - degree Taylor polynomial (P_4(x)) of (f(x)) about (x = 1), we have (P_4(x)=f(1)+f^{\prime}(1)(x - 1)+\frac{f^{\prime\prime}(1)}{2!}(x - 1)^{2}+\frac{f^{\prime\prime\prime}(1)}{3!}(x - 1)^{3}+\frac{f^{(4)}(1)}{4!}(x - 1)^{4}).

Step2: Find the fourth - derivative of (f)

We are given (f^{\prime\prime\prime}(x)=2x^{-3}). Differentiate (f^{\prime\prime\prime}(x)) with respect to (x) using the power rule ((x^n)^\prime=nx^{n - 1}). So (f^{(4)}(x)=\frac{d}{dx}(2x^{-3})=- 6x^{-4}).

Step3: Evaluate the fourth - derivative at (x = 1)

Substitute (x = 1) into (f^{(4)}(x)). We get (f^{(4)}(1)=-6\times1^{-4}=-6).

Step4: Find the coefficient of ((x - 1)^4)

The coefficient of ((x - 1)^4) in the Taylor polynomial is (\frac{f^{(4)}(1)}{4!}). Since (4!=4\times3\times2\times1 = 24) and (f^{(4)}(1)=-6), then (\frac{f^{(4)}(1)}{4!}=\frac{-6}{24}=-\frac{1}{4}).

Answer:

A. (-\frac{1}{4})