Using calculus, how would you prove that $$y={{2\,x}\over{\left(x^3+1\right)}} $$ is not a 1-1 function?
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$\lim\limits_{x\to-\infty}\frac{2x}{x^3+1}=0$, $\lim\limits_{x\to-1_-}\frac{2x}{x^3+1}=+\infty$, but eg. $f(1)>0$.
user2345215
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Could you please let me know how this contradicts the injectivity of $f$. – Semsem Mar 09 '14 at 10:35
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@Semsem The function is continuous on $(-\infty,-1)$. – user2345215 Mar 09 '14 at 10:37
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For example $f(x)=\frac{4}{x-5}+5$ satisfies the same conditions and still 1-1 – Semsem Mar 09 '14 at 10:37
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@Semsem: It doesn't. – user2345215 Mar 09 '14 at 10:45
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Now i understand what you have done, i thought that the limits when $x\to \pm\infty$,Cheers – Semsem Mar 09 '14 at 10:52
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Or even more simple, but using the same point as in the answer of user2345215, $f(x)=1$ is equivalent to $$ 0=x^3-2x+1=(x^2+x-1)(x-1)=((x+0.5)^2-1.25)(x-1) $$ and from the last term one sees that there are three real solutions.
Lutz Lehmann
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In fact, it is even more trivial: if $y = 1$, then $2x = x^3 + 1$, or equivalently, $(x-1)(x^2+x-1) = 0$. The quadratic factor has a positive discriminant, so it admits two distinct real-valued roots, hence the function is not one-to-one, as it has three distinct solutions to $y = 1$.
But, using calculus, we can also compute $$y' = \frac{2(1-2x^3)}{(x^3+1)^2},$$ which obviously has a critical point at $x = 2^{-1/3}$. The second derivative is $$y'' = \frac{12x^2(x^3-2)}{(x^3+1)^3},$$ which is negative for $x = 2^{-1/3}$, so this critical point is a local extremum, and so the function cannot be one-to-one.
heropup
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There are several good answers, but you asked for one using calculus, so: by setting the derivative to zero, show there's a maximum, and conclude the function's not one-one.
Gerry Myerson
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