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Given are the following two functions:

$$g(z) = \left(z-2\right)\left(2+z\left(z-2\right)\right)$$


$$h(z) = 2\left(z-1\right)^{2}\ln\left(z-1\right),$$

where $z>2$.

I would like to show that these functions intersect only once (for $z>2$, clearly they also intersect at $z=2$). Graphically this seems to hold, but I would like to show it analytically as well. It's easy to show that both functions are strictly increasing and strictly convex. However, this does still allows for more than one intersection. I hope someone can help me to show this as I have been struggling with it for days. Thanks for your help!

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Should indeed be z. – Abe Doe Aug 2 '12 at 9:06
up vote 3 down vote accepted

Let $f(x)=g(x)-h(x)$. We know that $f(2)=0$, and want to show that it has it has exactly one other zero in $[2,\infty]$. We have $$\begin{align} f(x)&=(x-2) ((x-2) x+2)-2 (x-1)^2 \log (x-1),\\ f'(x)&=3 x^2-10 x-4 (x-1) \log (x-1)+8,\\ f''(x)&=6 x-4 \log (x-1)-14,\\ f'''(x)&=6-\frac{4}{x-1}. \end{align}$$

One can verify that $f'(2)=0$ and $f''(2)<0$, so $f(2+\epsilon)<0$ for sufficiently small $\epsilon$, and that $f(x) > 0$ for sufficiently large $x$. So there is at least one other zero.

Suppose there were two zeroes apart from $x=2$, say $x_1$ and $x_2$. Then by Rolle's theorem, there would be at least two zeroes of $f'(x)$: one between $2$ and $x_1$, and one between $x_1$ and $x_2$. But $f'(x)$ is convex for $x\ge2$, and since it starts at $f'(2)=0$ with negative derivative, it cannot cross zero more than once in $x>2$.

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That's exactly what I wanted to show. I am eternally in your debt! – Abe Doe Aug 2 '12 at 9:08

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