# 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.

• Isn't the function $x\log(x^2-1)$ increasing? – crystal_math Jan 19 at 20:21

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 $$(1,\infty)$$.

• isnt this missing the $(x^2-1)^{x-1}$ at the start – user845712 Jan 20 at 18:58
• @EliThompson: It's positive so we need not worry about it. – The73SuperBug Jan 21 at 0:00

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.

$$g(x)=\left(x^2-1\right)^x$$ $$g(x)$$ 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 $$x$$.