Find the range of $f \colon \mathbb{R} \to \mathbb{R}$, $f(x) = 2x^3 + 3x^2 − 4$. Is $f$ injective? Is $f$ surjective?

Question

A function $f \colon \mathbb{R} \to \mathbb{R}$ is defined by $f(x) = 2x^3 + 3x^2 − 4$.

Find the range of $f$. Is $f$ one–to–one (injective)? Is $f$ onto (surjective)? Is $f$ a bijection? Give reasons for all your answers.

I recently attempted to solve a problem of this type, but was completely unsure of what to do. The only thing I remember is that one or more of these properties could be checked by taking either the first or second derivatives (again, can't remember).

I would greatly appreciate it if people could please take the time to explain how one would go about solving this type of problem.

Answer 1

We have that f(x) is continuos and

• $\lim_{x\to \infty} f(x)=\infty$
• $\lim_{x\to -\infty} f(x)=-\infty$

therefore by IVT $f(x)$ is surjective.

Note also that

$$f'(x)=6x^2+6x=6x(x+1) \quad f''(x)=12x+6$$

therefore $x=0$ is a local minimum and $x=-1$ is alocal maximum, thus $f(x)$ is not injective.

Answer 2

I'm not sure you necessarily need to go to the derivatives. Consider the following:

"Injective" - does it take the same value for different inputs? If so, it's not injective; if not, it is.

"Surjective" - since it's defined as a function from the reals onto the reals, does it reach every real? If so, it's surjective, if not, not.

Of course, it's bijective iff it's both injective and surjective.

Hint: an odd polynomial from the reals onto the reals will always be surjective.

Answer 3

• A third degree polynomial function is always surjective (onto $\mathbb{R}$) because: $$\lim_{x\to\pm\infty}\left(ax^3+bx^2+cx+d\right)=\pm\infty \quad\quad(a > 0)$$ or $$\lim_{x\to\pm\infty}\left(ax^3+bx^2+cx+d\right)=\mp\infty \quad\quad(a < 0)$$ and polynomials are continuous. Your $f(x)$ is such a polynomial, hence the range is $\mathbb{R}$.

• It is only injective if it is monotonic, i.e. if the derivative has a constant sign. In your case the derivative is $f'(x)=6x(x+1)$ so it does not have a constant sign and therefore $f$ is not monotonic, hence not injective.

• A function is bijective if (and only if) it is injective and surjective, so yours is not.

Answer 4

We have that f(x) is continuos and

limx→∞f(x)=∞ limx→−∞f(x)=−∞ therefore by IVT f(x) is surjective.

Note also that

f′(x)=6x2+6x=6x(x+1)f′′(x)=12x+6 therefore x=0 is a local minimum and x=−1 is alocal maximum, thus f(x) is not injective.