1. **Problem statement:** Find all local extrema, the global maximum, and the global minimum of the functions: c) $$f(x) = e^{-\frac{x^2}{2}}$$ d) $$f(x) = \frac{x^2}{x-2}$$ --- 2. **Function c: $$f(x) = e^{-\frac{x^2}{2}}$$** - Domain: $$\mathbb{R}$$ (all real numbers) - Find critical points by computing the derivative: $$f'(x) = e^{-\frac{x^2}{2}} \cdot \left(-x\right) = -x e^{-\frac{x^2}{2}}$$ - Set derivative to zero: $$-x e^{-\frac{x^2}{2}} = 0 \implies x = 0$$ - Second derivative test: $$f''(x) = \frac{d}{dx}(-x e^{-\frac{x^2}{2}}) = -e^{-\frac{x^2}{2}} + x^2 e^{-\frac{x^2}{2}} = e^{-\frac{x^2}{2}}(x^2 - 1)$$ - Evaluate at $$x=0$$: $$f''(0) = e^0 (0 - 1) = -1 < 0$$ - Since $$f''(0) < 0$$, $$x=0$$ is a local maximum. - Behavior at infinity: $$\lim_{x \to \pm \infty} e^{-\frac{x^2}{2}} = 0$$ - So the global maximum is at $$x=0$$ with value: $$f(0) = e^0 = 1$$ - The global minimum is approached as $$x \to \pm \infty$$ with value 0 (not attained). --- 3. **Function d: $$f(x) = \frac{x^2}{x-2}$$** - Domain: $$x \in [-5,5] \setminus \{2\}$$ (exclude $$x=2$$ where denominator is zero) - Find critical points by computing the derivative using quotient rule: $$f'(x) = \frac{(2x)(x-2) - x^2(1)}{(x-2)^2} = \frac{2x^2 - 4x - x^2}{(x-2)^2} = \frac{x^2 - 4x}{(x-2)^2}$$ - Set numerator to zero: $$x^2 - 4x = 0 \implies x(x-4) = 0 \implies x=0 \text{ or } x=4$$ - Check if these points are in domain: Both 0 and 4 are in $$[-5,5]$$ and not equal to 2, so valid. - Evaluate $$f(x)$$ at critical points and endpoints: $$f(0) = \frac{0}{0-2} = 0$$ $$f(4) = \frac{16}{4-2} = \frac{16}{2} = 8$$ $$f(-5) = \frac{25}{-5-2} = \frac{25}{-7} \approx -3.571$$ $$f(5) = \frac{25}{5-2} = \frac{25}{3} \approx 8.333$$ - Check behavior near the vertical asymptote $$x=2$$: As $$x \to 2^-$$, denominator $$\to 0^-$$, numerator $$\to 4$$, so $$f(x) \to -\infty$$. As $$x \to 2^+$$, denominator $$\to 0^+$$, numerator $$\to 4$$, so $$f(x) \to +\infty$$. - Determine local extrema by testing intervals or second derivative: Second derivative is complicated, so test values: - For $$x=1$$ (between 0 and 2): $$f'(1) = \frac{1 - 4}{(1-2)^2} = \frac{-3}{1} = -3 < 0$$ (decreasing) - For $$x=3$$ (between 2 and 4): $$f'(3) = \frac{9 - 12}{(3-2)^2} = \frac{-3}{1} = -3 < 0$$ (decreasing) - For $$x=5$$ (greater than 4): $$f'(5) = \frac{25 - 20}{(5-2)^2} = \frac{5}{9} > 0$$ (increasing) - So at $$x=4$$, derivative changes from negative to positive, indicating a local minimum. - At $$x=0$$, derivative changes from positive (left) to negative (right), indicating a local maximum. - Summary: Local maximum at $$x=0$$ with $$f(0)=0$$. Local minimum at $$x=4$$ with $$f(4)=8$$. - Global maximum on $$[-5,5]$$ is at $$x=5$$ with $$f(5) \approx 8.333$$. - Global minimum is unbounded near $$x=2^-$$ tending to $$-\infty$$. --- **Final answers:** c) Local and global maximum at $$x=0$$ with $$f(0)=1$$. Global minimum approaches 0 as $$x \to \pm \infty$$. d) Local maximum at $$x=0$$ with $$f(0)=0$$. Local minimum at $$x=4$$ with $$f(4)=8$$. Global maximum at $$x=5$$ with $$f(5) \approx 8.333$$. Global minimum tends to $$-\infty$$ near $$x=2^-$$ (not attained).