Given the function:
[tex]f(x)=sinx+1[/tex]
Let's find the minimum and maximum values over the interval [0, 2π].
Let's first find the derivative of the function:
[tex]f^{\prime}(x)=cosx[/tex]
Now set the derivative to 0 and solve for x:
[tex]\begin{gathered} cosx=0 \\ \\ \text{ Take the inverse cosine of both sides:} \\ x=cos^{-1}(0) \\ \\ x=\frac{\pi}{2} \end{gathered}[/tex]
The cosine function is positive in quadrants I and IV, to find the reference angle(minimum), subtract the first solution from 2π:
[tex]\begin{gathered} x=2\pi-\frac{\pi}{2} \\ \\ x=\frac{2(2\pi)-\pi}{2} \\ \\ x=\frac{4\pi-\pi}{2} \\ \\ x=\frac{3\pi}{2} \end{gathered}[/tex]
Plug in the values in the function to determine the minimum and maximum:
[tex]\begin{gathered} f(\frac{\pi}{2})=sin(\frac{\pi}{2})+1=1+1=2 \\ \\ \\ f(\frac{3\pi}{2})=sin(\frac{3\pi}{2})+1=-1+1=0 \end{gathered}[/tex]
Therefore, we have the following:
Minimum occurs at: x = 3π/2
Maximum occurs at: x = π/2
ANSWER:
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