# Determining whether $y=\sqrt{x^3+x^2+x+1}$ is one-to-one.

How can I determine that $y=\sqrt{x^3+x^2+x+1}$ is a one-to-one function?

I attempted to use the method in answer to this question...

$\sqrt{x^3+x^2+x+1}=^?\sqrt{y^3+y^2+y+1}$

$x^3+x^2+x+1=^?y^3+y^2+y+1$

$x^3+x^2+x=^?y^3+y^2+y$

I don't see a way to continue the proof to $y=x$ any further.

In general, how can I determine whether some function is one-to-one?

If you knew what derivatives were, then you could easily do this question. With that knowledge, I knew the function was one-one.

But to come to your question, you are on the right track, but a little more hard work is required: $$x^3+x^2+x=y^3+y^2+y \implies (x^3-y^3) + (x^2-y^2) + (x-y) = 0$$

As it turns out, we can factorize the above expression: $$(x^3-y^3) + (x^2-y^2) + (x-y) = (x-y)(1 + x+y+x^2+xy+y^2) =0$$ Now, all we need to show is that $1 + x+y+x^2+xy+y^2$ is a strictly positive quantity, so that $x-y=0$ is forced by the previous statement.

Now, we complete various squares, and we can rewrite the expression: \begin{split} 1 + x+y+x^2+xy+y^2 & = \left(\frac{x+y}{2}\right)^2 + \frac{3}{4}x^2 + \frac{3}{4}y^2 + x+y+1 \\ & = \left(\frac{x+y}{2}\right)^2 + \frac{(3x+2)^2}{12} + \frac{9y^2+12y+8}{12} \\ & = \left(\frac{x+y}{2}\right)^2 + \frac{(3x+2)^2}{12} + \frac{(3y+2)^2}{12} + \frac{1}{3} \end{split}

Which means that $1+x+y+x^2+xy+y^2 \geq \frac 13 >0$.

Hence, from the previous equality, we get $x-y=0$, so that $x=y$. Hence, the function given is one-one.

If you do not know derivatives, then little tricks like completing squares and factorisations are immensely helpful, but in general you are going to struggle if I give you some function with lots of $\ln$s and sines and cosines. So you'll have to wait until calculus to get a more comprehensive answer to this question.

• Why does that expression need to be strictly positive? – Fine Man Nov 8 '16 at 7:17
• The square of any quantity is non-negative, hence $(3x+2)^2$, $\left(\frac{x+y}{2}\right)^2$ and $(3y+2)^2$ are all greater than or equal to zero. Furthermore, if you divide by a positive quantity, the sign does not change, so $\frac{(3x+2)^2}{12}$ etc. all remain positive. What remains is $\frac{1}{3}$, so the whole thing is greater than $\frac 13$. – астон вілла олоф мэллбэрг Nov 8 '16 at 7:29
• Sorry, I was too ambiguous. What I meant was: why does $1+x+y+x^2+xy+y^2$ need to be strictly positive? What purpose does the fact that it ought to be positive serve? – Fine Man Nov 8 '16 at 22:50
• See, suppose we know that for two quantities $a$ and $b$, $ab=0$, but $a>0$. Then, because the product is zero, $b=0$ is forced. Now, in the above example, $a = 1+x+y+x^2+xy+y^2$ and $b=x-y$. $ab=0$ was derived by us, and we showed that $a>0$, so $b=0$ must happen. – астон вілла олоф мэллбэрг Nov 9 '16 at 9:36
• But what I'm not getting is why can't $a$ be negative (i.e. as long as $a \neq 0$)? Wouldn't $b$ still be forced to be $0$? – Fine Man Nov 11 '16 at 4:21

Suppose that $0 \le x < y$, then $x^2<y^2$ and $x^3<y^3$. Hence

$$x^3+x^2+x < y^3+y^2+y$$

I know you included the tag (algebra-precalculus), but just for those who stumble across this and happen to know calculus, proving a function is $1-1$ is equivalent to showing that the derivative is always positive or always negative. This is equivalent to having the original function be always increasing or always decreasing, which is a sufficient condition for being $1-1$.

In this case the derivative is a quadratic, namely $3x^2+2x+1$. Since we already have the tools of calculus we can find the minimum of this function. We note that the minimum will occur when the derivative of the quadratic is zero, i.e. $6x+2=0 \implies x=-\frac 13$. Evaluating our quadratic at this point yields $\frac 23$, and further noting that the quadratic function grows to $+\infty$ without bounds we are done.