
What is the domain and range of $y=-\sqrt{1-x}?$
Answer
528.6k+ views
Hint: We know that if we define a function $f:A\to B$ from a non-empty set $A$ to a non-empty set $B,$ then the domain of the function is $A$ and the range of the function is a subset of $B$ that consists of all the images of elements in the domain under the function.
Complete step by step solution:
Let us consider the given function $y=-\sqrt{1-x}.$
We are asked to find the domain and the range of the given function.
We will first tend to the domain of the function.
We know that the domain of a function is the set of elements from which the values mapped to their images.
And we also know that there will not be any element in the domain that does not map to an image in the range set.
We should remember that no element can have more than one image, though two elements can have the same image.
We know that the function is defined only when the term under the square root is greater than or equal to zero.
So, we will get $1-x\ge 0.$
So, when we transpose $x,$ we will get $1\ge x.$
Therefore, the domain is $x\le 1.$
Now, let us find the range of the function.
As we know, the range of the function is the set of images of the elements in the domain.
When we have $x\le 1,$ we will get $1-x\ge 0.$ So, we can say that the function is defined and the function attains all the non-positive values.
Therefore, the range is $y\le 0.$
Hence the domain of the function is the set of all values $x\le 1$ and the range of the function is the set of values $y\le 0.$
Note: We know that the set $B$ where $f:A\to B$ is called the codomain of the function. If all the elements of the codomain have corresponding preimages in the domain, we say that the range is equal to the codomain. Or if there are elements that do not have preimages, then the range is a subset of the codomain.
Complete step by step solution:
Let us consider the given function $y=-\sqrt{1-x}.$
We are asked to find the domain and the range of the given function.
We will first tend to the domain of the function.
We know that the domain of a function is the set of elements from which the values mapped to their images.
And we also know that there will not be any element in the domain that does not map to an image in the range set.
We should remember that no element can have more than one image, though two elements can have the same image.
We know that the function is defined only when the term under the square root is greater than or equal to zero.
So, we will get $1-x\ge 0.$
So, when we transpose $x,$ we will get $1\ge x.$
Therefore, the domain is $x\le 1.$
Now, let us find the range of the function.
As we know, the range of the function is the set of images of the elements in the domain.
When we have $x\le 1,$ we will get $1-x\ge 0.$ So, we can say that the function is defined and the function attains all the non-positive values.
Therefore, the range is $y\le 0.$
Hence the domain of the function is the set of all values $x\le 1$ and the range of the function is the set of values $y\le 0.$
Note: We know that the set $B$ where $f:A\to B$ is called the codomain of the function. If all the elements of the codomain have corresponding preimages in the domain, we say that the range is equal to the codomain. Or if there are elements that do not have preimages, then the range is a subset of the codomain.
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