Let $f(x)=\sin x$. Determine the $x$-value(s) where the function has a maximum or minimum value on $[0,2 \pi)$. To enter $\pi$, type Pi (with a capital P) The fields below accept a list of numbers or formulas separated by semicolons (e.g. $2 ; 4 ; 6$ or $x+1 ; x-1$ ). The order of the lists do not matter. On $[0,2 \pi)$, the maximum value(s) of the function occur(s) at what $x$-value(s)? \[ x= \] On $[0,2 \pi)$, the minimum value(s) of the function occur(s) at what $x$-value(s)? \[ x= \]
Solution
Step 1 ：The function \(f(x) = \sin x\) is a periodic function with period \(2\pi\).
Step 2 ：It reaches its maximum value of 1 at \(x = \frac{\pi}{2} + 2n\pi\) and its minimum value of -1 at \(x = \frac{3\pi}{2} + 2n\pi\), where \(n\) is an integer.
Step 3 ：However, since we are only interested in the interval \([0, 2\pi)\), we only need to consider the values of \(n\) that make \(x\) fall within this interval.
Step 4 ：For the maximum value, this happens when \(n = 0\), giving \(x = \frac{\pi}{2}\).
Step 5 ：For the minimum value, this also happens when \(n = 0\), giving \(x = \frac{3\pi}{2}\).
Step 6 ：Final Answer: The maximum value of the function \(f(x) = \sin x\) on \([0,2 \pi)\) occurs at \(x = \frac{\pi}{2}\) and the minimum value occurs at \(x = \frac{3\pi}{2}\).
Step 7 ：So, the final answer is \(x_{\text{max}} = \boxed{\frac{\pi}{2}}\) and \(x_{\text{min}} = \boxed{\frac{3\pi}{2}}\)
