Determine if the function is onto, if not find an element of the codomain that is not in the range. I have no clue where to begin with this. I appreciate any help.
g:R-->[1,infinity) given by g(x)=x^2+1
Read the introduction in wikipedia explaining the difference between range and codomain.
For example, if
g:R->R g(x)=x^2, then the codomain is all of R, but the range (image) is [0,∞)
For this problem, g is onto, since both the codomain and the image are [1,∞). There is no element of [1,∞) such that g(x) cannot be found there.
To determine if a function is onto (or surjective), we need to check if every element in the codomain has a corresponding element in the range. In other words, we need to determine if for every "y" in the codomain, there exists an "x" in the domain such that g(x) = y.
In this case, the function g: R --> [1, infinity) is defined as g(x) = x^2 + 1.
To show that g is onto, we need to demonstrate that for any given y in [1, infinity), we can find an x in the domain such that g(x) = y.
To do this, let's first consider an arbitrary value of y, say y = 5.
We need to solve the equation g(x) = 5, which translates to x^2 + 1 = 5.
Subtracting 1 from both sides, we have x^2 = 4.
Taking the square root of both sides, we get x = ±2.
Therefore, for y = 5, we have two possible values of x: x = 2 and x = -2.
By finding solutions to g(x) = y for any y in [1, infinity), we can see that the function g is onto.
If a function is not onto, we need to find an element in the codomain that is not in the range.
In this case, since the function g is onto, there is no element in the codomain [1, infinity) that is not in the range.