In Exercises 7 – 12, find a formula for the \(n^{th}\) term of the Taylor series of \(f(x)\), centered at \(c\), by finding the coefficients of the first few powers of \(x\) and looking for a paƩern. (The formulas for several of these are found in Key Idea 8.8.1; show work verifying these formula.)
\(f(x)=\frac{1}{x}\\f'(x)=\frac{-1}{x^2}\\f''(x)=\frac{2}{x^3}\\f'''(x)=\frac{-6}{x^4}\\f''''(x)=\frac{24}{x^5}\\ \text{Let}\space a=1\\f(1)=1\\f'(1)=-1\\f''(1)=2\\f'''(1)=-6\\f''''(1)=24\\1+\frac{-1}{1!}(x-1)+\frac{2}{2!}(x-1)^2+\frac{-6}{3!}(x-1)^3+\frac{24}{4!}(x-1)^4\\1-(x-1)+(x-1)^2-(x-1)^3+(x-1)^4...\\=\sum\limits_{n=0}^{\infty}(-1)^n(x-1)^n\\\)