Solve the equation : $ e^x(x^2-1)>1$ I came across a question in a quiz for my son as he asked  for help.
Well I went through classic way  and got a mixture of exponential and quadratic expression, I derived to look for the extremum then moved to second derivative. But I felt I was locked in the same problem.
so we have $ P(x)=  x^2e^x -e^x -1 >0$
solving for P'(x)=0 to look for extremums, we have $e^x(x^2 +2x-1)= 0 \rightarrow  x_{1,2}=\frac{-2\mp \sqrt8}{2} $
So $x_{1,2}=\frac{-2\mp \sqrt8}{2} $  are P'(x) roots
The probem is that the graph plotting gives differents roots.
As I wonder how shoudl I proceeed, I  had thought to Newton method.
But is it the only way to do so .
Thanks for any help.
 A: If you look for the zero of function
$$f(x)=e^x(x^2-1)-1$$ a simple inspection shows that the solution is greater than $1$.
Make a series expansion
$$f(x)=-1+2 e (x-1)+3 e (x-1)^2+2 e (x-1)^3+\frac{5}{6} e (x-1)^4+O\left((x-1)^5\right)$$ Now, series reversion
$$x=1+\frac{f(x)+1}{2 e}-\frac{3 (f(x)+1)^2}{8 e^2}+\frac{7 (f(x)+1)^3}{16 e^3}-\frac{235 (f(x)+1)^4}{384 e^4}+O\left((f(x)+1)^5\right)$$ SInce you want $f(x)=0$, an estimate is
$$x=1+\frac{1}{2 e}-\frac{3 }{8 e^2}+\frac{7 }{16 e^3}-\frac{235 }{384 e^4}=1.14376$$ while the solution given by Newton method is $1.14776$.
A: First we calculate the derivative of $f(x)=e^{x}(x^{2}-1)$. It is clear that the inequality can be true only for $x>1$ or $x<-1$.
$f'(x)=e^{x}(x^{2}+2x-1)$. The roots of the binomial are $-1+\sqrt{2}$ and $-1-\sqrt{2}$.
Hence $f$ is increasing for $x>-1+\sqrt{2}$ and clearly for $x>1$.
All we have to do is to find for which value we have $f(x)=1$ which turns out to be $x=1.14776$.
Thus for $x>1.14776$ the inequality is true.
Now we will prove that for $x<-1 , f(x)<1.$ Set $y=-x$ and then $y>1$.
We shall prove that $e^{-y}(y^{2}-1)<1$ which is equivalent to:
$e^{y}>\,y^{2}-1$. By using elementary calculus we can prove that this holds for $y>1$.
Thus, the inequality is true only for $x>1.14776..$
