select the correct answer from each drop - down menu. the graph of the function (f(x)=\frac{5}{4}sin(x)+1)…

Explanation:

Step1: Recall the range of sine function

The range of $y = \sin(x)$ is $[- 1,1]$. For the function $y=\frac{5}{4}\sin(x)+1$, we first consider the transformation of the sine - function.

Step2: Find the minimum value

When $\sin(x)=-1$, $y=\frac{5}{4}\times(-1)+1=-\frac{5}{4}+1=-\frac{1}{4}$.

Step3: Find the maximum value

When $\sin(x) = 1$, $y=\frac{5}{4}\times1 + 1=\frac{5}{4}+1=\frac{9}{4}$.

Step4: Determine the range on the given interval

We need to check the values of the function $y = \frac{5}{4}\sin(x)+1$ on the interval $(0,\frac{\pi}{2})$. The sine function $y=\sin(x)$ is increasing on the interval $(0,\frac{\pi}{2})$. When $x = 0$, $y=\frac{5}{4}\sin(0)+1=1$; when $x=\frac{\pi}{2}$, $y=\frac{5}{4}\sin(\frac{\pi}{2})+1=\frac{5}{4}+1=\frac{9}{4}$. So on the interval $(0,\frac{\pi}{2})$, the minimum value of the function is $1$, the maximum value of the function is $\frac{9}{4}$, and the range of the function is $(1,\frac{9}{4})$.

Answer:

The minimum value of the function on the interval $(0,\frac{\pi}{2})$ is $1$, the maximum value of the function is $\frac{9}{4}$, and the range of the function is $(1,\frac{9}{4})$.