1. **State the problem:** Analyze the function $$f(x) = e^{2x} - 3e^x - 2x$$ including its limits as $$x \to \pm \infty$$, derivative, critical points, variation, and intersections with the line $$y = -2x$$. 2. **Limits at infinity:** - As $$x \to -\infty$$, $$e^x \to 0$$, so $$e^{2x} = (e^x)^2 \to 0$$ and $$-3e^x \to 0$$. The term $$-2x \to +\infty$$ because $$x$$ is large negative and multiplied by $$-2$$. - Therefore, $$\lim_{x \to -\infty} f(x) = 0 - 0 - 2x = +\infty$$. - As $$x \to +\infty$$, $$e^x$$ grows faster than any polynomial, so $$e^{2x}$$ dominates. - Hence, $$\lim_{x \to +\infty} f(x) = +\infty$$. 3. **Derivative:** Use the derivative rules: $$f'(x) = \frac{d}{dx}(e^{2x}) - \frac{d}{dx}(3e^x) - \frac{d}{dx}(2x) = 2e^{2x} - 3e^x - 2$$ 4. **Critical points:** Set $$f'(x) = 0$$: $$2e^{2x} - 3e^x - 2 = 0$$ Let $$t = e^x > 0$$, then: $$2t^2 - 3t - 2 = 0$$ Solve quadratic: $$t = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4}$$ - $$t_1 = \frac{8}{4} = 2$$ - $$t_2 = \frac{-2}{4} = -0.5$$ (discard negative since $$t>0$$) So $$e^x = 2 \Rightarrow x = \ln 2$$ is the only critical point. 5. **Sign of derivative:** - For $$x < \ln 2$$, test $$x=0$$: $$f'(0) = 2 - 3 - 2 = -3 < 0$$, so $$f' < 0$$. - For $$x > \ln 2$$, test $$x=1$$: $$f'(1) = 2e^{2} - 3e - 2 > 0$$ (since $$e^2 \approx 7.39$$, $$e \approx 2.72$$), so $$f' > 0$$. 6. **Variation:** - $$f$$ decreases on $$(-\infty, \ln 2)$$ and increases on $$(\ln 2, +\infty)$$. - At $$x = \ln 2$$, $$f$$ has a minimum. 7. **Intersection with $$y = -2x$$:** Solve $$f(x) = -2x$$: $$e^{2x} - 3e^x - 2x = -2x \Rightarrow e^{2x} - 3e^x = 0$$ Factor: $$e^x(e^x - 3) = 0$$ Since $$e^x > 0$$, we have: $$e^x - 3 = 0 \Rightarrow e^x = 3 \Rightarrow x = \ln 3$$ 8. **Summary:** - $$\lim_{x \to -\infty} f(x) = +\infty$$ - $$\lim_{x \to +\infty} f(x) = +\infty$$ - Minimum at $$x = \ln 2$$ - Intersects $$y = -2x$$ at $$x = \ln 3$$ Final answer: The function decreases until $$x=\ln 2$$, then increases, with a minimum there, and crosses the line $$y=-2x$$ at $$x=\ln 3$$.