Whether or not is is a function depends on how the mapping is defined. If, for example the mapping is f(x, y) = x, where the coordinates of points in 2-d space are mapped to their abscissa or g(x, y) = y, where the coordinates of points in 2-d space are mapped to their ordinates then they are functions.
That depends on the specific function.
A function is a mapping from one set to another. It may be many-to-one or one-to-one. The first of these sets is the domain and the second set is the range. Thus, for each value x in the domain, the function allocates the value f(x) which is a value in the range. For example, if the function is f(x) = x^2 and the domain is the integers in the interval [-2, 2], then the range is the set [0, 1, 4].
x y -3 2 -1 6 1 -2 3 5
yes y=x Like 2=2
Domain (input or 'x' values): -∞ < x < ∞.Range (output or 'y' values): -2 ≤ y ≤ 2.
The domain of the function 1/2x is {0, 2, 4}. What is the range of the function?
That depends on the specific function.
The answer depends on the domain. If the domain is the whole of the real numbers, the range in y ≥ 1. However, you can choose to have the domain as [1, 2] in which case the range will be [2, 5]. If you choose another domain you will get another range.
A function is a mapping from one set to another. It may be many-to-one or one-to-one. The first of these sets is the domain and the second set is the range. Thus, for each value x in the domain, the function allocates the value f(x) which is a value in the range. For example, if the function is f(x) = x^2 and the domain is the integers in the interval [-2, 2], then the range is the set [0, 1, 4].
x y -3 2 -1 6 1 -2 3 5
The range is {-7, 1, 9, 17}.
yes y=x Like 2=2
Yes. Typical example: y = x2. To avoid comparing infinite sets, restrict the function to integers between -3 and +3. Domain = -3, -2 , ... , 2 , 3. So |Domain| = 7 Range = 0, 1, 4, 9 so |Range| = 4 You have a function that is many-to-one. One consequence is that, without redefining its domain, the function cannot have an inverse.
Domain (input or 'x' values): -∞ < x < ∞.Range (output or 'y' values): -2 ≤ y ≤ 2.
The function ( f(x) = x^2 + 1 ) is a quadratic function. The domain is all real numbers, expressed as ( (-\infty, \infty) ), since you can input any real number for ( x ). The range is ( [1, \infty) ) because the minimum value of the function occurs at ( x = 0 ), where ( f(0) = 1 ), and the function increases without bound as ( x ) moves away from zero.
To find the range of the function ( f(x) = -10x ) for the given domain (-4, -2, 0, 2, 4), we can evaluate the function at each point in the domain. For ( x = -4 ), ( f(-4) = 40 ) For ( x = -2 ), ( f(-2) = 20 ) For ( x = 0 ), ( f(0) = 0 ) For ( x = 2 ), ( f(2) = -20 ) For ( x = 4 ), ( f(4) = -40 ) Thus, the range of the function is ([-40, 40]).
The domain of a function pertains to all the x values The range of a function pertains to all the y values So domain and range do not get confused, this can be easily remembered by the order of the how the first letter of the word appears in the English alphabet. d, domain, goes before r, range x goes before y domain = x values range = y values ill try to add to the previous writer. previously, he wrote what the domain and range are for easier functions, but not how to determine them. more generally, what the domain is, is what you can put into a function, which in simpler cases, is jus x. to find what you can put in, it helps to find what you cant put in for x, meaning, where is the graph of the function discontinuous. for example, if we look at the function f(x)=1/(1-x) if we put 1 in for x, then the denominator goes to zero and the function is discontinuous at that x value, therefore 1 will not be included in the domain, but everything else will be included since there are no other disconinuities. the domain will end up looking like this- (-infinity,1), (1,infinity). now for the range, all you have to do is find what you can get out of the function from what you can put in, which can usually be done by putting the values you see for the domain in. putting negative infinity in for x in f(x)=1/(1-x) you get zero and putting one in you get infinty. putting it together you get (-infinity,0), (0,infinity) for your range. p.s. as i stated before, more generally, your domain is more so what you put into your function, so it is not always x, for example, in the case of a function of 2 variables such as f(x,y), what you can put in for both x and y will make up your domain, not just x, and y will most certainly not be your range, rather it will be the values of f(x,y).