$\displaystyle x^2+e^{-x}=8$
Is there an analytic solution? If not, why so?
$\displaystyle e^xx^2+1=8e^x\Rightarrow{e}^x=\frac{-1}{x^2-8}$ ?
All this tells me is that $\displaystyle |x|>\sqrt8$
$\displaystyle x^2+e^{-x}=8$
Is there an analytic solution? If not, why so?
$\displaystyle e^xx^2+1=8e^x\Rightarrow{e}^x=\frac{-1}{x^2-8}$ ?
All this tells me is that $\displaystyle |x|>\sqrt8$
I didn't check your calculations but from the last equation you know that $\displaystyle -\sqrt{8} < x < \sqrt{8}$:
$\displaystyle e^x$ is positive
Therefore the denominator of the fraction must be negative
EDIT:
Probably you have noticed that your last equation can't be used to solve it for x because you have the variable on both sides of the equation.
I "solved" (using Newton's method) the equation:
$\displaystyle x^2+e^{-x}-8=0$
and I got 2 real approximate results: $\displaystyle x_1 \approx 2.817847...~\vee~x_2\approx -1.65826...$