Let $f$ be a function defined as $$f:=\sqrt[3]{x^{3}+bx^{2}}, b\in \mathbb{R}$$ for all$$x\leq 0$$
In order for a function ta have an inverse, it must be both injective and surjective(hope I spelled that one correctly). It is easy to show that $f$ is an injective function, meaning that $x_{1}= x_{2}$ implies $f\left ( x_{1} \right )= f\left ( x_{2} \right )$, and vice-versa. Proving surjecitvity means proving that for every element of the codomain there exists at least one corresponding element in the domain. Or in other words, the image of the domain is the whole codomain.But how do I find the set of numbers which go into the codomain, and how to prove surjectivity for this particular case? After not being able to complete the second step, I treid to solve for the inverse function $$y= \sqrt[3]{x^{3}+bx^{2}} ,$$ $$x=\sqrt[3]{y^{3}+by^{2}}$$ $$y=\sqrt{\frac{x^{3}}{y+b}}$$ This is how far I've gotten.
My intuition tells me that I have made an error. Feel free to improve this question.
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$\begingroup$The cube root function is injective, and the composition of injective functions is injective. So your function $f$ is injective if and only if $x^3+bx^2$ is injective. Since that is continuously differentiable, that will be so iff the derivative of $x^3+bx^2$ never switches sign in its domain (avoiding having both positive and negative values: zero at one point is fine).
So find the values of $b$ for which the derivative of $x^3+bx^2$ does and does not switch sign, and that will tell you the values of $b$ for which $f$ is not and is injective. (Keeping in mind the restriction $x\le 0$, of course.)
Since $\lim_{x\to -\infty}(x^3+bx^2)=-\infty$ and its value at zero is zero, it is surjective on $(-\infty,0)$, and thus so is your function $f$. You need only worry about it being injective.
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