Prove function $f$ is not onto

Question

Let $f: \mathbb R \to \mathbb R$ defined by $$f(x) = 2x^{2}+x.$$ Prove that $f$ is not onto.

The problem is, I think it IS onto because for every $y$ I try to create there's an $x$ that can make it through the function. Why am I wrong?

Answer 1

Hint: note that the codomain is $\mathbb{R}$. (For example: what is the preimage of $y=-42$?)

Answer 2

Hint:

• Try to find the minimal value that this function can attain, perhaps by completing the square or calculus.

• From the answer in the previous part, you should be able to construct a number with no preimage.

Answer 3

HINT

Let consider the limit for $x\to \pm \infty$and observe that $f(x)$ is continuous.

Answer 4

Completing the square $2x^2+x=2(x+1/4)^2-1/8$, observe that a square is always $\geq0$. Therefore, values $<-1/8$ cannot be attained.

Answer 5

Let $m \in \mathbb R$

If $f(x) = m$ would mean $2x^2 + x = m$ so $2x^2 + x - m = 0$ so $x =\frac {-1\pm \sqrt {1 + 8m}}{4}$.

Can that always happen? Are there any $m$ were that can't happen?

Well, it can't happen if $1 + 8m < 0$ or $m < -\frac 18$.

And that makes sense. The $x$ "part" of $2x^2 + x$ dips down into the negatives and can be as negative as we want, but the $2x^2$ "part" bobs right back up into the positives and if $2x^2 > |x|$ the "positive force" of $2x^2$ overwhelms the "negative force" of $x$ and we just can't sink to those negative values. And $2x^2$ "overpowers" $x$ fairly early.