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Let $A = \mathbb{R}\setminus[−1,1]$. Let $g : A\to\mathbb{R}$ be defined by $g(x)=(x^2-1)^x$ for all $x\in A$.

Prove that $g(x)=(x^2-1)^x$ is increasing on $(1,\infty)$

I have currently attempted to prove this by showing $g'(x)\geq 0 $ for all $ x\in A $ which gives $g'(x)= (x^2-1)^{x-1}(2x^2+(x^2-1)ln(x^2-1))$ which should be greater than or equal to $0$ however I am unsure how to show this or whether I have gone about this the right way.

Any Help will be grateful.

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  • $\begingroup$ Isn't the function $x\log(x^2-1)$ increasing? $\endgroup$ – crystal_math Jan 19 at 20:21
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Follow what you left off, $2x^2 + (x^2-1)\ln(x^2-1)= 2(x^2-1)+(x^2-1)\ln(x^2-1)+2= 2t+t\ln t + 2=h(t), t = x^2-1>0$. Computing $h'(t) = 3+\ln t= 0 \iff t=e^{-3}$. Observe that $h''(t) = \dfrac{1}{t} > 0, t > 0$. Thus $h_{\text{min}} = h(e^{-3})= 2e^{-3}+2+-3e^{-3}= 2-e^{-3} > 0$. By calculus's $2$nd second derivative test, it shows that $h(t) > 0$, and thus your $g'(x) > 0$ and the function is increasing over $(1,\infty)$.

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  • $\begingroup$ isnt this missing the $(x^2-1)^{x-1}$ at the start $\endgroup$ – user845712 Jan 20 at 18:58
  • $\begingroup$ @EliThompson: It's positive so we need not worry about it. $\endgroup$ – The73SuperBug Jan 21 at 0:00
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A. For b>1 x^b is increasing. B. For a>1 a^x is increasing.

For any x_1 < x_2 where both are bigger than root 2:

We get from A above:

[(x_1)^2 - 1]^x_1 < [(x_2)^2 - 1]^x_1

Then we get from B above:

[(x_2)^2 - 1]^x_1 < [(x_2)^2 - 1]^x_2

putting them together the function is increasing when bigger than root 2.

Only very close to 1 do we need further analysis

d/dx((x^2 - 1)^x) = (x^2 - 1)^x ([(2 x^2)/(x^2 - 1)] + log(x^2 - 1))

So the front part of the right hand side of the above equation is always positive for x>1,so we only need to make sure that [(2 x^2)/(x^2 - 1)] + log(x^2 - 1) is always positive for x>1.

So we want to show 2x^2 + (x^2-1)log(x^2-1) >0 for x>1. Let U = x^2 -1, U>0 for x>1, so it suffices to show 2 + 2U + Ulog(U) > 0 for U>0.

d/dx(x log(x)) = log(x) + 1

Ulog(U) takes a minimum point at U= 1/e. So Ulog(U) > -1/e for U>0. That proves the inequality:

2x^2 + (x^2-1)log(x^2-1) >0 for x>1.

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$$g(x)=\left(x^2-1\right)^x$$ $g(x)$ is increasing if $g'(x)>0$

$$g'(x)=\left(x^2-1\right)^x \left(\frac{2 x^2}{x^2-1}+\log \left(x^2-1\right)\right)$$ $g'(x)>0$ if $$\frac{2 x^2}{x^2-1}+\log \left(x^2-1\right)>0$$ to simplify set $x^2=z+1$ $$\frac{2z+2}{z}+\log z>0\to h(z)=2+\frac{2}{z}+\log z>0$$ This is true for any $z$, because minimum value of $h(z)$ is positive

Indeed $h'(z)=-\frac{2}{z^2}+\frac{1}{z}=0\to z=2$

$h''(z)=\frac{4}{z^3}-\frac{1}{z^2}$

$z=2$ is a minimum because $h''(2)=\frac{4}{8}-\frac{1}{4}>0$ (second derivative test)

$h(2)=3+\log 2>0$, therefore $g'(x)>0$ for any $x$.

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