Determine algebraically if the function $f(x) = x^2 + 4x$ is onto or not

Question

Determine algebraically if the function $f(x) = x^2 + 4x$ is onto or not

I do not know how to approach this question, how should I go about it. How should I factor $x^2 + 4x$?

Answer 1

What they want you to do is to examine whether there can ever be an $x$ such that $x^2+4x=-10$ (i.e. solve that equation). Because if the function is onto, then there should be such an $x$. (Note: there is nothing really special about $-10$ here, other than being smaller than $-4$.)

Answer 2

You should know a method that symbolically allows you to calculate the solutions of the equation $$x^2+4x=y$$ in terms of $y$, and which also allows you to decide when such solutions exist in the real numbers. You should use it here.

Answer 3

To say that $f$ is onto means that for all real numbers $b$, the equation $f(x) = b$ has a solution $x$. Thus, we ask does $x^2 + 4x = b$ always have solutions? The answer is given by the quadratic formula: $$ x = \frac{-4 \pm \sqrt{16 +4b}}{2}. $$ If $16 + 4b \ge 0$, then there is a solution $x$. However, we can see that some values of $b$ make $16 + 4b < 0$. That means that $x$ would have to be complex for $f(x) = b$. In other words, $f$ is not onto since for $b \lt -4$, there do not exist real numbers $x$ such that $f(x) = b$.

If you wanted to vet this with analytic geometry. The graph of $f$ is an upward-facing parabola. Parabolas have vertices, and the graph of an upward-facing parabola cannot "reach" points below the vertex, so our answer makes sense with what we know from analytic geometry.

Answer 4

If the function is onto (onto what? the reals?) then for every $y$ there is an $x$ such that $f(x) =x^2 + 4x = y$.

So $x^2 + 4x -y = 0$. So $x = \frac{-4 \pm \sqrt{16 + 4y}}{2}$.

But if $y < -4$ then $16 + 4y < 0$ and there is no such real $x$. So there are no $x$ so that $f(x) = y; y < -4$.

So $f: \mathbb R \rightarrow \mathbb R$ is not onto.

HOWEVER

If $f: \mathbb R \rightarrow [-4, \infty)$ then $f$ is onto.

Or if $f: \mathbb C \rightarrow C$ then $f$ is onto.

Answer 5

You don't need to factor the polynomial. I assume that the question is if $f:\mathbb{R}\to\mathbb{R}$ given by $f(x) = x^2 +4x$ is onto.

This is not the case. There are a couple of ways to see this

1. From the graph you can see that you can draw a horizontal line that does not intersect the graph.
2. Maybe also from a graph you can guess a number (like $-10$ as suggested in another answer) $a$ such that $x^2 +4x = a$ does not have any solution. All you are trying to do is to find this $a$ such that the discriminant $4^2 - 4\cdot 1 \cdot (-a)$ is negative. That is, you need to find an $a$ such that $16 + 4a$ is negative. This shouldn't be too hard.
3. You can also appeal to the fact that all polynomials are continuous and so since the domain is the set of all real numbers and $\lim_{x\to \pm \infty} x^2 + 4x = \infty$, the function will attain a minimal value. Hence there is a smallest value and so $x^2 + 4x$ can't be smaller than that. Hence the function is not onto.

Answer 6

After completing the square you can rewrite the given function as $$f(x) = (x^2+4x+4)-4 = (x+2)^2-4.$$ So you immediately see that $f(x)\ge -4$ for every real number $x$. This means that there is no $x$ such that $f(x)=-5$, the function is not surjective (onto).