91Ó°ÊÓ

First, we need to find the first few derivatives of the given function, \(f(x) = \cos x\). Let's do it up to order 2. First derivative: \(f'(x) = -\sin x\) Second derivative: \(f''(x) = -\cos x\)

Now, evaluate these derivatives at x = 0: \(f(0) = \cos 0 = 1\) \(f'(0) = -\sin 0 = 0\) \(f''(0) = -\cos 0 = -1\)

Use the evaluated derivatives to construct the Taylor polynomials: 0th-order Taylor polynomial: \(P_0(x) = f(0) = 1\) 1st-order Taylor polynomial: \(P_1(x) = f(0) + f'(0)(x-0) = 1\) 2nd-order Taylor polynomial: \(P_2(x) = f(0) + f'(0)(x-0) + \frac{f''(0)(x-0)^2}{2!} = 1 - \frac{x^2}{2}\)

Now, graph the function \(f(x) = \cos x\) and the Taylor polynomials \(P_0(x), P_1(x),\) and \(P_2(x)\) together. It's important to observe how the approximation gets better as the degree\( n\) of the Taylor polynomial increases. When graphing them, you'll notice the following: - The 0th-order Taylor polynomial \(P_0(x) = 1\) is just a horizontal line that approximates \(\cos x\) at \(x = 0\). - The 1st-order Taylor polynomial \(P_1(x) = 1\) also approximates \(\cos x\) at \(x = 0\) but doesn't provide any extra information. - The 2nd-order Taylor polynomial \(P_2(x) = 1 - \frac{x^2}{2}\) provides a better approximation near \(x = 0\), capturing the concave shape of the function \(\cos x\) in a vicinity of the point \(x=0\). In conclusion, we found the nth-order Taylor polynomials for the function \(f(x) = \cos x\) centered at x = 0, for \(n=0,1,\) and 2, and we graphed them along with the function. Observe how the approximation improves as the order \(n\) of the Taylor polynomial increases.