graph the function ( y = 3sinleft(\frac{1}{3}x\right) ). show at least two cycles. use the graph to…
graph the function ( y = 3sinleft(\frac{1}{3}x\right) ). show at least two cycles. use the graph to determine the domain and range of the function.\nuse the graphing tool to graph the equation. type pi to insert ( pi ) as needed.\nuse the graph to determine the domain of ( y = 3sinleft(\frac{1}{3}x\right) ).\n(type your answer in interval notation. use integers or fractions for any numbers in the expression.)\nuse the graph to determine the range of ( y = 3sinleft(\frac{1}{3}x\right) ).\n(type your answer in interval notation. use integers or fractions for any numbers in the expression.)
Answer
Explanation:
Step1: Recall the domain of sine function
The general form of a sine function is (y = A\sin(Bx - C)+D). For the function (y=\sin(u)), the domain of (u) (in our case (u = \frac{1}{3}x)) is all real numbers. Since there are no restrictions on the value of (x) for which (\frac{1}{3}x) is defined (because for any real - number (x), the operation of multiplying (x) by (\frac{1}{3}) is valid), the domain of (y = 3\sin(\frac{1}{3}x)) is the set of all real numbers. In interval notation, the set of all real numbers is ((-\infty,\infty)).
Step2: Recall the range of sine function
The range of the basic sine function (y=\sin(t)) is ([- 1,1]). For the function (y = A\sin(Bx - C)+D), the range is given by ([D - |A|,D + |A|]). In the function (y = 3\sin(\frac{1}{3}x)), we have (A = 3), (B=\frac{1}{3}), (C = 0), and (D = 0). We know that (-1\leqslant\sin(\frac{1}{3}x)\leqslant1). Multiply each part of the inequality by (3): [ \begin{align*} 3\times(-1)&\leqslant3\sin(\frac{1}{3}x)\leqslant3\times1\ -3&\leqslant y\leqslant3 \end{align*} ] In interval notation, the range is ([-3,3]).
Answer:
- Domain: ((-\infty,\infty))
- Range: ([-3,3])