I have a cut off function
$$0 \leq \chi(x) \leq 1, \ \chi(x) = \begin{cases} 0, \ |x\pm 1|\leq 1/8 \\ 1, \ |x\pm1| \geq 1/4 \end{cases}$$ and I would like to estimate the function $$f(x) := \chi(x)xe^{- t(1-x^2)^2} \lesssim Ce^{-t\omega}$$
Where $t>0$ is fixed and $C,\omega$ are positive constants which need not be sharp and are independent of $x$. This should be straight forward, but I am not sure how to do this.
First note that $\chi$ makes the problem unnecessarily difficult ($|f|$ is larger if one would leave out $\chi$) $$ |f(x)| \leq |x| \exp(-t(1-x^2)^2) \,. $$ Therefor, consider $g(x)=: x \exp(-t(1-x^2)^2)$.
Find extrema of $g$ \begin{align} 0 &\overset{!}{=} g'(x) = \exp(-t(1-x^2)^2) \left[ 1-4tx^2 (x^2-1) \right] \\ \iff \quad 0 &= x^4 - x^2 - \frac{1}{4t} \\ \implies \quad x^2 &= \frac{1}{2} \left( 1+ \sqrt{1+\frac{1}{t}} \right) \\ \implies \quad x_0 &:= \sqrt{\frac{1}{2} \left( 1+ \sqrt{1+\frac{1}{t}} \right) } \,. \end{align}
Given the extrema $\pm x_0$ it directly follows how to choose the constants \begin{align} C &:= x_0 \\ \omega &:= (1-x_0^2)^2 = \frac{1}{4} \left(2+\frac{1}{t}-2\sqrt{1+\frac{1}{t}} \right) \,. \end{align}