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$\newline$The function $\newline$$f$ has a Taylor series about $x=1$ that converges to )$f(x)$ for all $\newline$$x$ in the interval of convergence. It is known that $f(1)=1$,$f'(1)=-\frac{1}{2}$, and the $\newline$$n$th derivative of $f$ at $x=1$ is given by $f^{(n)}(1)=(-1)^{n}\frac{(n-1)!}{2^{n}}$ for$x=1$$0$.$\newline$(a) Write the first four nonzero terms and the general term of the Taylor series for$f$ about $x=1$.$\newline$(b) The Taylor series for $f$ about $x=1$ has a radius of convergence of $x=2$. Find the interval of convergence. Show the work that leads to your answer.$\newline$(c) The Taylor series for $f$ about $x=1$ can be used to represent $\newline$ $f(1.2)$ as an alternating series. Use the first three nonzero terms of the alternating series to approximate $f(1.2)$.$\newline$(d) Show that the approximation found in part (c) is within $0.001$ of the exact value of $f(1.2)$.