91Ó°ÊÓ

We have been given a function $y=5-3\sin \left(2x\right)$.

We have to graph this function and determine the domain and the range of the function using the graph.

On comparing, we get: $$\begin{gathered} \left|A\right|=3 \\ T=\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }} \\ \mathrm{Ó\neg }=2;T=\mathrm{Ï€} \end{gathered}$$

The graph lies between $3$and $-3$.

One cycle starts from $x=0$and ends at $x=\mathrm{Ï€}$.

Five key points have x-coordinates as:$0,\frac{\mathrm{Ï€}}{4},\mathrm{Ï€},\frac{3\mathrm{Ï€}}{2},2\mathrm{Ï€}$

For the second cycles, x-coordinates of key points will be:$\frac{5\mathrm{Ï€}}{2},3\mathrm{Ï€},\frac{7\mathrm{Ï€}}{2},4\mathrm{Ï€}$

x and y coordinates for the first cycles are:$(0,0),\left(\frac{\mathrm{π}}{4},−3\right),\left(\frac{\mathrm{π}}{2},0\right),\left(\frac{3\mathrm{π}}{4},3\right)(\mathrm{π},0)$

x and y coordinates for the second cycles are:$\left(\frac{5\mathrm{π}}{4},−3\right),\left(\frac{3\mathrm{π}}{2},0\right)\left(\frac{7\mathrm{π}}{4},3\right),(2\mathrm{π},0)$

x and y coordinates for the first cycles will become: $(0,5),\left(\frac{\mathrm{Ï€}}{4},2\right),\left(\frac{\mathrm{Ï€}}{2},5\right),\left(\frac{3\mathrm{Ï€}}{4},8\right),(\mathrm{Ï€},5)$

x and y coordinates for the second cycles will become:

$\left(\frac{5\mathrm{Ï€}}{4},2\right),\left(\frac{3\mathrm{Ï€}}{2},5\right)\left(\frac{7\mathrm{Ï€}}{4},8\right),(2\mathrm{Ï€},5)$

The domain of this function can be determine through the graph, since it continues indefinitely to the left and to right, the domain is the set of all real numbers. Thus, the domain is

$\left\{x\right|-∞<x<∞\}$

Through the graph the range of the function consists of all real numbers from $2$to $8$, inclusive Hence, the range is

role="math" localid="1646385406595" $\left\{y\right|2≤y≤8\}$