Prove that $g(x)=(x^2−1)^x$ is increasing on $(1,+\infty)$ 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.
 A: 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 $(1,\infty)$.
A: 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.
A: $$g(x)=\left(x^2-1\right)^x$$
$g(x)$ 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 $x$.
