question 4 of 10 for the function y = -2 + 5 sin(π/12(x - 2)), what is the maximum value? answer here

Explanation:

Step1: Recall sine - function range

The range of the sine function $y = \sin(u)$ is $[- 1,1]$, i.e., $-1\leqslant\sin(u)\leqslant1$. In the given function $y=-2 + 5\sin\left(\frac{\pi}{12}(x - 2)\right)$, let $u=\frac{\pi}{12}(x - 2)$.

Step2: Find the maximum of the sine - part

To find the maximum value of $y=-2 + 5\sin\left(\frac{\pi}{12}(x - 2)\right)$, we need to find the maximum value of $\sin\left(\frac{\pi}{12}(x - 2)\right)$. The maximum value of $\sin\left(\frac{\pi}{12}(x - 2)\right)$ is 1.

Step3: Calculate the maximum of the function

Substitute $\sin\left(\frac{\pi}{12}(x - 2)\right)=1$ into the function $y=-2 + 5\sin\left(\frac{\pi}{12}(x - 2)\right)$. Then $y=-2+5\times1$. $y=-2 + 5=3$.

Answer:

3