Question

How do you find the domain and range of the given function $y=-3\sin \left( \dfrac{1}{2} \right)x$?

Answer 1

Hint: We start solving the problem by making use of the fact that the sine function $a\sin \left( bx \right)$ is defined at every real value to find the domain of the given function. We then recall the fact that the range of the sine function $\sin \left( bx \right)$ is $\left[ -1,1 \right]$ to get the range of $y=\sin \left( \dfrac{1}{2} \right)x$. We then multiply this range with –3 to get the required range of the given function.

Complete step by step answer:
According to the problem, we are asked to find the domain and range of the given function $y=-3\sin \left( \dfrac{1}{2} \right)x$.
We have given the function $y=-3\sin \left( \dfrac{1}{2} \right)x$.
We know that the sine function $a\sin \left( bx \right)$ is defined at every real value. This tells us that the domain of a sine function is $\mathbb{R}$, which is the domain of the given function.
So, the domain of the given function $y=-3\sin \left( \dfrac{1}{2} \right)x$ is $\mathbb{R}$.
Now, let us find the range of the given function. We know that the range of the sine function $\sin \left( bx \right)$ is $\left[ -1,1 \right]$.
So, the range of the function $y=\sin \left( \dfrac{1}{2} \right)x$ as $\left[ -1,1 \right]$.
Since the $y=\sin \left( \dfrac{1}{2} \right)x$ is multiplied with –3, every term present in its range is multiplied with 3 to get the range of the function $y=-3\sin \left( \dfrac{1}{2} \right)x$.

So, the range of the function $y=-3\sin \left( \dfrac{1}{2} \right)x$ is $-3\times \left[ -1,1 \right]=\left[ -3,3 \right]$.

Note: We should keep in mind that the range must be represented in the form $\left[ a,b \right]$, where $a < b$, otherwise the answer is incorrect. We should keep in mind that the standard trigonometric functions have the domain as $\mathbb{R}$ while solving this type of problems. Similarly, we can expect problems to find the domain and range of the function $y=5\cos \left( \dfrac{1}{{{x}^{2}}-1} \right)$.