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 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.
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.
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.