Determine the concavity of y = 3 + sin x on [0, 2π\piπ]. | Basic Mathematics 2080

Basic Mathematics20805 marksNumerical: concavity of functionConcavity and inflection points

Determine the concavity of y = 3 + sin x on [0, 2π\piπ]. [5]

Function: $y = 3 + \sin x$ - Interval: $[0, 2\pi]$ $y' = \cos x$ $y'' = -\sin x$ Concave up requires $y'' 0$ : $-\sin x 0 \implies \sin x < 0$ On $[0, 2\pi]$ , $\sin x < 0$ for $x \in (\pi, 2\pi)$ . Concave down requires $y'' < 0$ : $-\sin x < 0 \implies \sin x 0$ On $[0, 2\pi]$ , $\sin x 0$ ...

Function: $y = 3 + \sin x$ - Interval: $[0, 2\pi]$ $$y' = \cos x$$ $$y'' = -\sin x$$ Concave up requires $y'' 0$: $$-\sin x 0 \implies \sin x < 0$$ On $[0, 2\pi]$, $\sin x < 0$ for $x \in (\pi, 2\pi)$. Concave down requires $y'' < 0$: $$-\sin x < 0 \implies \sin x 0$$ On $[0, 2\pi]$, $\sin x 0$...