1. Problem 31: Find relative asymptotes and study variation of $f(x) = x e^x$ 2. To find asymptotes, check behavior as $x \to \pm \infty$: - As $x \to \infty$, $e^x$ dominates, so $f(x) = x e^x \to \infty$. - As $x \to -\infty$, $e^x \to 0$, but $x$ goes to $-\infty$, so $f(x) \to 0$. No vertical or horizontal asymptotes. 3. Study monotonicity by differentiating: $$f'(x) = e^x + x e^x = e^x (1 + x)$$ 4. Critical points where $f'(x) = 0$: $$e^x (1 + x) = 0 \Rightarrow 1 + x = 0 \Rightarrow x = -1$$ 5. Analyze sign of $f'(x)$: - For $x < -1$, $1 + x < 0$, so $f'(x) < 0$, decreasing. - For $x > -1$, $1 + x > 0$, so $f'(x) > 0$, increasing. 6. At $x = -1$, local minimum. Calculate $f(-1) = (-1) e^{-1} = -\frac{1}{e}$. 7. Problem 32: $f(x) = 2^{2x}$. 8. Rewrite with properties: $$f(x) = (2^2)^x = 4^x$$ 9. Check asymptotes: - As $x \to \infty$, $f(x) \to \infty$. - As $x \to -\infty$, $f(x) \to 0$. Horizontal asymptote: $y = 0$ as $x \to -\infty$. No vertical asymptotes. 10. Derivative: $$f'(x) = 4^x \ln 4 > 0$$ $f$ strictly increasing. 11. Problem 33: $f(x) = \frac{1}{x} e^x$ 12. Domain: $x \neq 0$. 13. Check asymptotes: - As $x \to 0^+$, $f(x) \to \infty$. - As $x \to 0^-$, $f(x) \to -\infty$. Vertical asymptote at $x=0$. 14. As $x \to \infty$, $e^x / x \to \infty$, so no horizontal asymptote. 15. Derivative using quotient/product rule: $$f'(x) = \left(\frac{1}{x}\right)' e^x + \frac{1}{x} (e^x)' = \left(-\frac{1}{x^2}\right) e^x + \frac{1}{x} e^x = e^x \left(-\frac{1}{x^2} + \frac{1}{x}\right) = e^x \frac{x - 1}{x^2}$$ 16. Critical points where $f'(x)=0$: $$\frac{x-1}{x^2} = 0 \Rightarrow x = 1$$ 17. Sign of $f'(x)$ (for $x \neq 0$): - For $x < 0$, numerator $x-1 < 0$, denominator $x^2 > 0$, so $f'(x) < 0$, decreasing. - For $0 < x < 1$, numerator $x-1 < 0$, so $f'(x) < 0$, decreasing. - For $x > 1$, numerator $x-1 > 0$, $f'(x) > 0$, increasing. 18. At $x=1$, local minimum. Calculate $f(1) = 1 \cdot e^{1} = e$. 19. Problem 34: $f(x) = 2^{4 \ln x}$, $x > 0$ since $ ln$ defined for positive $x$. 20. Rewrite: $$f(x) = 2^{4 \ln x} = e^{\ln(2) \cdot 4 \ln x} = e^{4 \ln(2) \ln x} = (e^{\ln x})^{4 \ln 2} = x^{4 \ln 2}$$ 21. No vertical asymptotes; domain $x>0$. 22. For $x \to 0^+$, since $4 \ln 2 > 0$, $f(x) \to 0$. For $x \to \infty$, $f(x) \to \infty$. 23. Derivative: $$f'(x) = 4 \ln 2 \cdot x^{4 \ln 2 - 1}$$ Always positive for $x > 0$. 24. $f$ strictly increasing on $(0, \infty)$. Answer summary: - 31: No asymptotes, local minimum at $x=-1, f(-1)=-1/e$ - 32: Horizontal asymptote $y=0$ at $x\to-\infty$, strictly increasing - 33: Vertical asymptote at $x=0$, local minimum at $x=1, f(1)=e$ - 34: No asymptotes, strictly increasing for $x>0$