Answer
$P_{2}(x) = \pi^{2} - 2\pi(x-\pi) - \frac{\pi^{2} - 2}{2!}(x-\pi)^{2}$
Work Step by Step
Given: $f(x) = x^{2} cos(x)$
$c=\pi$
$P_{n} (x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^{2} + \frac{f'''(c)}{3!}(x-c)^{3} + ... \frac{f^{n}(c)}{n!}(x-c)^{n}$
$f(\pi) = (\pi)^{2} cos(\pi) = -\pi^{2}$
$f'(x) = 2xcos(x) - x^{2}sin(x) → f'(\pi) = 2\pi cos(\pi) - (\pi)^{2} sin(\pi) = -2\pi$
$f''(x) = -x(3sin(x) + xcos(x)) - x sin (x) + 2 cos (x) → f''(\pi) = -\pi(3 sin (\pi) + \pi cos (\pi)) - \pi sin (\pi) + 2 cos (\pi) = -\pi^{2} - 2$
Taylor polynomial for $n=2$ is
$P_{2}(x) = \pi^{2} - 2\pi(x-\pi) - \frac{\pi^{2} - 2}{2!}(x-\pi)^{2}$
