Locate critical points and classify local extrema

Differentiate to get $d'(x) = -\sin(x) - x$. Setting $d'(x) = 0$ gives $-\sin(x) - x = 0$, i.e. $\sin(x) = -x$. On $[-\pi,\pi]$, we check for solutions: at the endpoints, $\sin(\pm\pi) = 0$ while $-x = \mp\pi \neq 0$, so the endpoints are not solutions. By symmetry and the bounded range of sine, the only solution in the interior is $x = 0$.

To classify $x=0$, check the sign of $d'(x) = -\sin(x) - x$ nearby: for small $x>0$, $\sin(x) \approx x$, so $d'(x) \approx -2x < 0$; for small $x<0$, $d'(x) \approx -2x > 0$. So $d'$ changes from positive to negative as $x$ increases through $0$, meaning $d(x)$ increases then decreases — $x=0$ is a local maximum, with $d(0) = 1$.

From $d'(x) = -\sin(x) - x$, differentiate again to get $d''(x) = -\cos(x) - 1$. Since $\cos(x) \le 1$ for all $x$, we have $d''(x) = -\cos(x) - 1 \le 0$ for every $x$, with equality only when $\cos(x) = -1$, which occurs at $x = \pi$ and $x = -\pi$ (the boundary points of the domain).

Because $d''(x) \le 0$ throughout $[-\pi, \pi]$ and is strictly negative except at the endpoints $x=\pm\pi$, the function is concave down on the entire interval and never changes concavity in the interior. An inflection point requires $d''$ to change sign; since $d''$ does not change sign in the interior, there are no inflection points in the open interval $(-\pi, \pi)$.

Since the function is concave down across the whole interval, the single critical point at $x=0$ cannot be a local minimum. A function that is everywhere concave down bends downward like an upside-down bowl, so any critical point must be a peak, i.e., a local maximum. This matches the earlier finding from part (a) that $x=0$ is a local maximum with $d(0)=1$, confirming that the two analyses are consistent.