1. The problem is to analyze the function $$f(x)=\frac{e^{5x}}{x^3-1}$$. 2. First, note the function involves an exponential numerator and a cubic denominator. 3. The denominator is $$x^3-1 = (x-1)(x^2+x+1)$$. 4. The function has vertical asymptotes where the denominator is zero, i.e., at $$x=1$$ and the roots of $$x^2+x+1=0$$ which are complex (no real roots). 5. So there is only one real vertical asymptote at $$x=1$$. 6. The numerator $$e^{5x}$$ is always positive for all real $$x$$. 7. To find the horizontal or slant asymptotes, consider $$x \to \pm \infty$$. 8. As $$x \to +\infty$$, $$e^{5x}$$ grows much faster than any polynomial, so $$f(x) \to \infty$$. 9. As $$x \to -\infty$$, $$e^{5x} \to 0$$ exponentially, while the denominator tends to $$-\infty$$, so $$f(x) \to 0^-$$. 10. To find critical points, compute the derivative using quotient rule: $$f'(x) = \frac{(e^{5x})'(x^3-1) - e^{5x}(3x^2)}{(x^3-1)^2} = \frac{5e^{5x}(x^3-1) - 3x^2 e^{5x}}{(x^3-1)^2} = \frac{e^{5x}(5x^3 - 5 - 3x^2)}{(x^3-1)^2}$$ 11. Set numerator zero for critical points: $$5x^3 - 3x^2 - 5=0$$ 12. This cubic equation may be solved numerically for critical points. 13. This analysis describes the function behavior, asymptotes, and critical points. Final: The function $$f(x)=\frac{e^{5x}}{x^3-1}$$ has a vertical asymptote at $$x=1$$, grows exponentially positively as $$x \to \infty$$, tends to zero from below as $$x \to -\infty$$, and critical points are roots of $$5x^3 - 3x^2 - 5=0$$.