# What is the domain and range of f(x) = 5 sin^2 x ?

##### 1 Answer
Oct 21, 2017

D: {x | x ∈ ℝ}
R: {y | 0 ≤ y ≤ 5}

#### Explanation:

The domain is the most obvious to figure out given that we know that $f \left(x\right) = \sin \left(x\right)$ is a function which has no bounds on the x value. A square of the function multiplied by five will not change the domain. These operations do not limit the domain. Therefore, the domain will continue to be of all real numbers.

The range, on the other hand, is a bit trickier to figure out. We know that the maximum and minimum values of $y = \sin \left(x\right)$ is 1 and -1. Squaring the maximum (sin(x)=1 rarr sin^2(x)=1^2=1) will make the maximum possible value for $y$ equal to 1. The minimum value is now 0 since for $\sin \left(x\right) = 0$, ${\sin}^{2} \left(x\right) = {0}^{2} = 0$
There can be no negatives since the function was squared.

The actual function is 5 times ${\sin}^{2} \left(x\right)$ so 5 times the minimum value is still 0 since $5 \cdot 0 = 0$

5 times the maximum is $5 \cdot 1 = 5$

So the final minimum value for y is 0 and final maximum value for y is 5.

Therefore

D: {x | x ∈ ℝ}
y: {y | 0 ≤ y ≤ 5}