You should remember that the image of the function if given by any element y from range of function, such that the equation `f(x) = y` has a solution, such that:

`(x^2 + 3x + 1)/(x^2 + 1) = y `

`x^2 + 3x + 1 = y(x^2 + 1) => x^2 + 3x + 1 - yx^2 - y = 0`

You need to consider the condition that the quadratic equation `x^2(1 - y) + 3x + 1 - y = 0` has solutions, hence, you need to use its discriminant `Delta = b^2 - 4ac` , such that:

`EE` `x_(1,2) in R if Delta = b^2 - 4ac >= 0`

Identifying the coefficients `a,b,c` yields:

`a = (1 - y), b = 3, c = 1 - y`

`x^2(1 - y) + 3x + 1 - y = 0`

`Delta = 9 - 4(1 - y)^2 >= 0`

You need to solve for y the inequality `9 - 4(1 - y)^2 >= 0` such that:

`9 - 4(1 - y)^2 = 0`

Converting the difference of squares into a product yields:

`(3 - 2(1 - y))(3 + 2(1 - y)) = 0 => (1 + 2y)(5 - 2y) = 0`

Using zero product rule yields:

`1 + 2y = 0 => y = -1/2`

`5 - 2y = 0 => y = 5/2`

You should notice that `9 - 4(1 - y)^2 >= 0` if `y in [-1/2,5/2]` , hence, the quadratic equation has solutions if `y in [-1/2,5/2]` .

**Hence, evaluating the image of the given function f(x) = y, yields **`y in [-1/2,5/2].`