![Where are the minimum and maximum values for f(x) = 2 sin x - 1 on the interval [0, 2π]?A. min:z = 0, 2 max:z =OB. min:z =max:z = 0,2플max:z = 0, 7, 2nC. min:z = class=](https://us-static.z-dn.net/files/dd6/308d329f5323b62e34562ae24a5946d9.png)
Recall that the maximum/minimum value of a function is the place where a function reaches its highest/lowest point.
We know that:
[tex]\begin{gathered} For\text{ }all\text{ }x\in[0,2\pi] \\ -1\leq\sin(x)\leq1. \end{gathered}[/tex]Multiplying the above result by 2 we get:
[tex]\begin{gathered} -1\times2\leq\sin(x)\times2\leq1\times2, \\ -2\leq2\sin(x)\leq2. \end{gathered}[/tex]Subtracting 1 from the above inequality we get:
[tex]\begin{gathered} -2-1\leq2\sin(x)-1\leq2-1, \\ -3\leq2\sin(x)-1\leq1. \end{gathered}[/tex]Therefore f(x) reaches a minimum at:
[tex]x=\frac{3\pi}{2}.[/tex]And f(x) reaches a maximum at:
[tex]x=\frac{\pi}{2}.[/tex]Answer: Option C.