1. **Stating the problem:** We analyze the function $$Y = e^x(e^x - 1)$$ by finding its domain, intercepts, sign study, limits, first and second derivatives. 2. **Domain:** The function is composed of exponential functions which are defined for all real numbers. \[ \text{Domain} = \mathbb{R} \] 3. **Intercepts:** - To find the y-intercept, evaluate at $$x=0$$: $$Y(0) = e^0(e^0 - 1) = 1 \times (1 - 1) = 0$$ - To find x-intercepts, solve $$Y=0$$: $$e^x(e^x - 1) = 0$$ Since $$e^x > 0$$ for all $$x$$, the only zero is when: $$e^x - 1 = 0 \Rightarrow e^x = 1 \Rightarrow x = 0$$ 4. **Sign study:** - For $$x < 0$$, $$e^x < 1$$ so $$e^x - 1 < 0$$, and since $$e^x > 0$$, the product is negative. - For $$x > 0$$, $$e^x - 1 > 0$$, so the product is positive. 5. **Limits:** - As $$x \to -\infty$$: $$e^x \to 0$$, so $$Y = e^x(e^x - 1) \approx 0 \times (0 - 1) = 0$$ More precisely, $$\lim_{x \to -\infty} Y = 0$$ - As $$x \to +\infty$$: $$e^x \to +\infty$$, so $$Y \approx e^x \times e^x = e^{2x} \to +\infty$$ 6. **First derivative:** Using product rule: $$Y = e^x(e^x - 1)$$ $$Y' = (e^x)'(e^x - 1) + e^x (e^x - 1)' = e^x(e^x - 1) + e^x(e^x) = e^x(e^x - 1 + e^x) = e^x(2e^x - 1)$$ 7. **Critical points:** Solve $$Y' = 0$$: $$e^x(2e^x - 1) = 0$$ Since $$e^x > 0$$, solve: $$2e^x - 1 = 0 \Rightarrow e^x = \frac{1}{2} \Rightarrow x = \ln\left(\frac{1}{2}\right) = -\ln 2$$ 8. **Second derivative:** Differentiate $$Y'$$: $$Y' = e^x(2e^x - 1)$$ $$Y'' = (e^x)'(2e^x - 1) + e^x (2e^x - 1)' = e^x(2e^x - 1) + e^x(2e^x) = e^x(2e^x - 1 + 2e^x) = e^x(4e^x - 1)$$ 9. **Concavity and inflection points:** Solve $$Y'' = 0$$: $$e^x(4e^x - 1) = 0$$ $$4e^x - 1 = 0 \Rightarrow e^x = \frac{1}{4} \Rightarrow x = -\ln 4$$ **Summary:** - Domain: $$\mathbb{R}$$ - Intercepts: at $$x=0$$ - Sign: negative for $$x < 0$$, positive for $$x > 0$$ - Limits: $$\lim_{x \to -\infty} Y = 0$$, $$\lim_{x \to +\infty} Y = +\infty$$ - Critical point at $$x = -\ln 2$$ - Inflection point at $$x = -\ln 4$$