SOLUTION 9: We are given the equation
$$ x e^x = x^2-1 \ \ \ \ \longrightarrow \ \ \ \ x e^x - x^2+1=0 $$
Let function
$$ f(x)= xe^x - x^2+1 \ \ \ \ and \ choose \ \ \ \ m=0 $$
This function is continuous for all values of $x$ since it is the SUM of continuous functions. We have the continuous polynomial function $1-x^2$. And the function $xe^x$ is continuous for all values of $x$ since it is the PRODUCT of $x$, a polynomial, and $e^x$, a well-known continuous function. We now need to search for an appropriate interval satisfying the assumptions of the Intermediate Value Theorem. By trial and error, we have that
$$ f(0)= (0)e^0 - (0)^2+1= 1 > 0 \ \ \ \ and \ \ \ \ f(-2)= (-2)e^{-2} - (-2)^2+1= \displaystyle{ -2 \over e^2 }-3 \approx -3.27 <0 $$
so that $$ f(-2) \approx -3.27 < m < 1 = f(0) $$
i.e., $m=0$ is between $ f(-2) $ and $ f(0) $.
The assumptions of the Intermediate Value Theorem have now been met, so we can conclude that there is some number $c$ in the interval $[-2, 0]$ which satisfies
$$ f(c)=m $$
i.e.,
$$ ce^c - c^2+1 =0 $$
and the equation is solvable.
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