Does there exists values of x for $f(x) = x^{2} e^{\frac{1}{(1-x^2)}}$ when $f(x)=1$ in $(-1,1)$

Question

Given $ \; \large{f(x)=x^{2} e^{\frac{1}{(1-x^2)}}} \; \; $ has solutions for $ f(x)=1 $ in $(-1,1)$. I think there are no solutions, but I am not sure how to prove it. Can anyone please help? Thanks in advance.

Answer 1

Hint

If you look for the roots of $$f(x)=x^{2} e^{\frac{1}{1-x^2}}-1=0$$ consider its derivative; after some minor simplifications, it writes $$f'(x)=2x\frac{ \left(x^4-x^2+1\right)}{\left(x^2-1\right)^2}e^{\frac{1}{1-x^2}}$$ which has only $x=0$ as real root and $f(0)=-1$. The second derivative test would show that this is a minimum. Since, $f(x)$ goes to $\infty$ when $x$ approaches $\pm 1$, then two roots in the interval .

Answer 2

Hint :

Study the function $ \; \large{f(x)=x^{2} e^{\frac{1}{(1-x^2)}}} \; \; $

Starting from $f(0)=0$, prove that the function is increassing in $x>0$ and decreassing in $x<0$