I need to prove that the following is not surjective. how do i do

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The function f: R -> R defined by f(x) = x^2 + 3x + 4 is not surjective. This conclusion is drawn from the observation that the minimum value of the function occurs at x = -1.5, yielding f(-1.5) = 1.75. Since the function is bounded below by 1.75, there are no real numbers x such that f(x) can equal any value less than 1.75. Additionally, the equation x^2 + 3x + 4 = 0 has no real solutions, confirming that the function does not cover all real numbers in its codomain.

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i need to prove that the following is not surjective. how do i do that?

let f:R->R be the function defined by f(x)=x^2 + 3x +4.
 
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You need to find at least one point in the codomain for which there is no mapping from the domain through f.

In other words, that function can only be surjective if we can choose some x to make f(x) equal to any value we like in R.

Clearly this function is bounded below by its minimum at x = -1.5,
which is f(-1.5) = 2.25 -4.5 + 4 = 1.75.

Observing that f has a minimum value of 1.75, we see that it is impossible to find any x in the real numbers such that f(x) is less than 1.75, thus f is not a surjective function.
 


donmkeys said:
i need to prove that the following is not surjective. how do i do that?

let f:R->R be the function defined by f(x)=x^2 + 3x +4.
The simplest thing to do is to note that the solutions to the equation [itex]x^2+ 3x+ 4= 0[/itex] are given by
[tex]\frac{-3\pm\sqrt{9- 4(1)(4)}}{2}= \frac{-3\pm\sqrt{-7}}{2}[/tex]
and so are not real numbers.
 

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