Consider the following function. \[ f(x)=\frac{2}{x}, \quad a=1, \quad n=2, \quad 0.6 \leq x \leq 1.4 \] (a) Approximate $f$ by a Taylor polynomial with degree $n$ at the number a. \[ T_{2}(x)= \] (b) Use Taylor's Inequality to estimate the accuracy of the approximation $f(x) \approx T_{n}(x)$ when $x$ lies in the given interval. (Round your answer to eight decimal places.) $\left|R_{2}(x)\right| \leq$

Step 1 ：We are given the function \(f(x) = \frac{2}{x}\) and we are asked to approximate this function by a Taylor polynomial of degree 2 at the point \(a = 1\).

Step 2 ：The Taylor series of a function about a point can be found using the formula: \[T_{n}(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} + \cdots\]

Step 3 ：First, we need to find the first and second derivatives of \(f(x)\). The first derivative of \(f(x) = \frac{2}{x}\) is \(f'(x) = -\frac{2}{x^2}\) and the second derivative is \(f''(x) = \frac{4}{x^3}\).

Step 4 ：Next, we evaluate these derivatives at \(x = 1\). We find that \(f(1) = 2\), \(f'(1) = -2\), and \(f''(1) = 4\).

Step 5 ：We substitute these values into the Taylor series formula to find the Taylor polynomial of degree 2: \[T_{2}(x) = f(1) + f'(1)(x-1) + \frac{f''(1)(x-1)^2}{2!} = 2 - 2(x - 1) + 2(x - 1)^2\].

Step 6 ：Simplifying this expression, we find that \[T_{2}(x) = -2x + 2(x - 1)^2 + 4\].

Step 7 ：Final Answer: The Taylor polynomial of degree 2 for the function \(f(x) = \frac{2}{x}\) at the point \(a = 1\) is \(\boxed{-2x + 2(x - 1)^2 + 4}\).