Answer
$ -6.78$
Work Step by Step
We are given that: $f(x) = x^{2} cos(x)$ when $c=\pi$
The polynomial at $x=a$ can be written as: $P_{n} (x) = f(c) + f'(c)(x-c) + \dfrac{f''(c)}{2!}(x-c)^{2} +.....+\dfrac{f^{n}(c)}{n!}(x-c)^{n}$
$f(\pi) = (\pi)^{2} \cos(\pi) = -\pi^{2} \\ f'(x) = 2xcos(x) - x^{2}sin(x) \implies f'(\pi) = 2\pi \cos(\pi) - (\pi)^{2} \sin \pi = -2\pi \\ f''(x) = -x(3 \sin(x) + x \cos(x)) - x \sin (x) + 2 \cos (x) \implies f''(\pi) = -\pi[3 \sin \pi + \pi \cos \pi] - \pi \sin (\pi) + 2 \cos \pi = -\pi^{2} - 2$
Thus, the Taylor polynomial for $n=2$ becomes:
$P_{2}(x) = \pi^{2} - 2\pi(x-\pi) - \dfrac{\pi^{2} - 2}{2!}(x-\pi)^{2}$
Now, replace $x$ with $\dfrac{7 \pi}{8}$ in the above equation to obtain:
$P_{2}(\dfrac{7 \pi}{8}) = \pi^{2} - 2\pi(\dfrac{7 \pi}{8}-\pi) - \dfrac{\pi^{2} - 2}{2!}(\dfrac{7 \pi}{8}-\pi)^{2} \approx -6.78$
