Which values of $(a,b)$ make a function bijective/surjective/injective.

Question

I am doing exam revision and there's this problem in one topic on functions:

Suppose,

$f$ : $\mathbb{R} \rightarrow \mathbb{R}$

Find $(a,b) \in \mathbb{R} \times \mathbb{R}$ : $f(x) = ax^3 -bx^2$ is,

• f is bijective.
• f is injective but not surjective.
• f is surjective but not injective.
• f is neither injective nor surjective.

The only method I've found helpful is essentially separating it into it's $x^2$ and $x^3$ components and then working out which values from there (either 1 or 0) yield these properties. I'm sure there is a more effective methods to find it out however.

Any advice/help is welcomed.

Answer 1

The beauty here is that we have to consider only the four cases of a or b being equal on not equal to zero.

1) the function is injective $\Leftrightarrow a\neq 0\ $ and $\ ax^3-bx^2$ has exactly one solution $\Leftrightarrow a \neq 0,\ b=0 $

2) function is surjective $\Leftrightarrow a \neq 0 $

Combining the cases, we obtain:

• bijective: $ \ a \neq 0,\ b=0$
• injective but not surjective : no way
• surjective but not injective : $ \ a \neq 0, \ b\neq 0 $
• no injective, no surjective : $ \ a = 0 $

Answer 2

If $a\ne0$, then the function is surjective (consider the limits at $\pm\infty$).

If $a=0$, the function is neither injective nor surjective (why?).

For $f$ being bijective, the derivative must…

Answer 3

** Hints**:

As $f$ is a differentiable function, ‘injective’ means ‘monotonic’, so itsderivative must have a constant sign, which isn't hard to test since it's a quadratic polynomial (at least if $a\ne 0$).

Also, by the Mean value theorem, a cubic polynomial function is always surjective from $\mathbf R$ to $\mathbf R$.