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I am struggling with this excercise:

I want to prove that the function $f: \mathbb{R} \to \mathbb{R}$, defined by $f(x)= x^3 + x^2 - 6x$, is surjective but not injective?

I personally would calculate some numbers and show that by these examples that this function cannot be injective. Is this way a correct way to prove this?

I appreciate your answer!

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user2051347
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4 Answers4

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Hint: Notice that $x^3+x^2-6x=x(x-2)(x-3)$ and then $\lim\limits_{x \to \pm \infty} f(x)= \pm \infty$.

Seirios
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Note that $\lim_{x\to \infty} f(x) = \infty$, while $\lim_{x \to -\infty}f(x) = -\infty$; surjectivity will then follow from the Intermediate Value Theorem.

On the other hand, $f$ has more than 1 root, so cannot be injective.

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$\lim\limits_{x\to+\infty}f(x) = +\infty$ and $\lim\limits_{x\to-\infty}f(x) = -\infty$ which implies that the function is surjective (intermediate value theorem).

To see that it is not injective, you can notice that $0$ is a root and that it has two other roots that are the roots of $x^2+x-6$, none of which are $0$. So you have two distinct numbers that have $0$ as image so it is not injective.

xavierm02
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Picking examples to show it's not injective is a fine way to do so. The simplest example to pick, two values $a,b$ such that $f(a) = 0$ and $f(b) = 0$, turns out to work fine.

If your initial guesses for which $c$ to use when trying to find two solutions to $f(x) = c$ don't work, you can appeal to your general approaches to understanding functions to try and gain insight as to which values would be useful to pick for $c$, such as graphing the function via calculator, or the techniques you've learned for plotting by hand.

The things that make it visually obvious that $f(x)$ is not injective, however, have a more direct translation to a mathematical argument than guessing a value for $c$. The same for surjectivity, incidentally:

See if you can prove the following two theorems:

If $f$ is an injective differentiable function, then $f'(x) \geq 0$ for all $x$, or $f'(x) \leq 0$ for all $x$.

If $f$ is a continuous function $\mathbf{R} \to \mathbf{R}$, then it is surjective if and only if it has no upper bound and no lower bound.

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