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.
To find the domain of the function ( f(x) = x^2 + 1 ), we identify the set of all possible input values for ( x ). Since this is a polynomial function, the domain is all real numbers, expressed as ( (-\infty, \infty) ). The range is determined by analyzing the output values; the minimum value of ( f(x) ) occurs at ( x = 0 ), giving ( f(0) = 1 ). Therefore, the range is ( [1, \infty) ).
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]).