How can I prove algebraically that the following function is not one-to-one

Question

$$f(x)= { x^3+1 \over x}$$

I know I have to start by setting $$f(a)=f(b)$$ and then try to show that

$$a = b$$

But I get stuck after clearing denominators and distributing.

Any help appreciated!

Answer 1

You don't have to find explicit $a$ and $b$ such that $f(a)=f(b)$, just show they exist. You can use continuity and the intermediate value theorem to do this.

If you divide it out you get $f(x)=x^2+\frac 1x$. Intuitively we know that the $x^2$ part is not one-to-one and when $x$ is large of either sign the $\frac 1x$ term is small. I would just note that $f(10)=\frac{1001}{10}, f(-10)=\frac {999}{10}, f(-11)=\frac {1330}{11} \gt \frac {1001}{10}$ so there is some value $x$ with $-11 \lt x \lt -10$ such that $f(x)=\frac {1001}{10}$ and the function is not one-to-one.

Answer 2

Let's determine the conditions which have to be satisfied by $x$ and $x'$ to have the same image: \begin{align} \frac{x^3+1}x=\frac{x'^3+1}{x'}&\iff(x^3+1)x'=(x'^3+1)x\iff x^3x'-x'^3x=x-x'\\ &\iff xx'(x^2-x'^2)=xx'(x-x')(x+x')=x-x'. \end{align} So, either $x=x'$, or $xx'(x+x')=1$.

This latter equation is a relation between the sum $s$ and the product $p$ of $x$ and $x'$. Conversely, if we know $s$ and $p$, $\;x$ and $x'$ are the roots of the quadratic equation $$u^2-su+p=0,$$ provided it has distinct real roots, i.e. provided $s^2>4p$.

Now, if $\;sp=1$, this condition is satisfied if $\;s^2>\frac4s$, i.e. $s^3>4$.

To sum this up, we can choose any $s>\sqrt[3]4$. Then the (real) roots of the equation $$u^2-su +\frac1s=0$$ have the same image by $f$.

Answer 3

A valid algebraic proof that $f$ is not injective could also go as follows:

Check whether there are $y$ for which the equation $$y = x^2 + \frac{1}{x}$$ has more than one solution. Rewriting the equation gives the cubic equation $$x^3 - yx + 1 = 0$$ So, now it is enough to find $y$ such that this equation has more than one real solution. Looking at the graph or a good algebraic intuition may hint you to $y=2$ as a good candidate: $$x^3 - 2x + 1 = (x-1)(x^2+x-1)= 0 \Rightarrow x_1 = 1,\; x_2 = -\frac{1}{2}\left( 1+\sqrt{5}\right),\;x_3 = -\frac{1}{2}\left( 1+\sqrt{5}\right)$$ And, indeed: $$f(x_1) = f(x_2) = f(x_3) = 2$$