Y = 1/X2 ==============Can it pass the line test? * * * * * That is not the inverse, but the reciprocal. Not the same thing! The inverse is y = sqrt(x). Onless the range is resticted, the mapping is one-to-many and so not a function.
The range depends on the domain. If the domain is the complex field, the range is also the whole of the complex field. If the domain is x = 0 then the range is 4.
x2+2x+1=y or y=x2 In this function the domain is x equals real values and the range is y equals all real values provided y is more than or equal to zero.
Y = X2 Is a parabolic function.
A function is a mapping from one set of numbers (domain) to another (range). The mapping need not be linear: it can be any mathematical function. That is, for every number in the domain the function provides a rule which allows you to calculate another number.If, then, you devise another function which is a mapping from the range of the first function to some other set, you have a function of a function.For example, suppose the first function, f, is "add 1" and the second function, g, is "square the number."Then the functiong of f = g[f(x)] = g[x+1] = [x+1]2 = x2 + 2x + 1however, note thatf of g = f[g(x)] = f[x2] = x2 + 1This illustrates that f of g is not the same as g of f.
No. It depends on how the range is defined.y = x2 is not onto R but can be made onto by changing the range to R0+.No. It depends on how the range is defined.y = x2 is not onto R but can be made onto by changing the range to R0+.No. It depends on how the range is defined.y = x2 is not onto R but can be made onto by changing the range to R0+.No. It depends on how the range is defined.y = x2 is not onto R but can be made onto by changing the range to R0+.
Y = 1/X2 ==============Can it pass the line test? * * * * * That is not the inverse, but the reciprocal. Not the same thing! The inverse is y = sqrt(x). Onless the range is resticted, the mapping is one-to-many and so not a function.
All the output values of a function are collectively called the "range" of that function. For example, consider the function x2. Any number squared will give a positive. Thus, the "range" of the function is positive numbers.
The answer, for y as a function of x, depends on the range of y. Over the real numbers, it is not a function because a function cannot be one-to-many. But it is always possible to define the domain and range in such a way that the mapping in not one-to-many.
The range depends on the domain. If the domain is the complex field, the range is also the whole of the complex field. If the domain is x = 0 then the range is 4.
x2+2x+1=y or y=x2 In this function the domain is x equals real values and the range is y equals all real values provided y is more than or equal to zero.
the function (x),sory I can`t use the sign of the function because it is not available. the function of (x)=4x+4 is one to one function assume function(x1)= function(x2) then 4(x1)+4 =4(x2)+4 4(x1)=4(x2) (x1)=(x2) hence,the function is one to one
No. x2+5x is a polynomial, an algebraic expression or a formula, but it is not a function. It could be used to help define a function. {(x,y) | y = x2 + 5x , x any real number} is a function
Y = X2 Is a parabolic function.
X2 9 can be described mathematically to be a quadratic function.
A function is a mapping from one set of numbers (domain) to another (range). The mapping need not be linear: it can be any mathematical function. That is, for every number in the domain the function provides a rule which allows you to calculate another number.If, then, you devise another function which is a mapping from the range of the first function to some other set, you have a function of a function.For example, suppose the first function, f, is "add 1" and the second function, g, is "square the number."Then the functiong of f = g[f(x)] = g[x+1] = [x+1]2 = x2 + 2x + 1however, note thatf of g = f[g(x)] = f[x2] = x2 + 1This illustrates that f of g is not the same as g of f.
If this is in the context of finding a root of an equation, the answer is to make some guesses. Find value x1 and x2 such that f(x1) and f(x2) have opposite signs. Then, provided that f is a continuous function over (x1, x2), the bisection method will find its root.