1. **Problem statement:** Find all local extrema, global maximum, and global minimum of each function on the domain $[-5,5]$. --- ### a) $f(x) = x^4 - 3x^3 + x^2 - 5$ 1. Find the derivative: $$f'(x) = 4x^3 - 9x^2 + 2x$$ 2. Set derivative to zero to find critical points: $$4x^3 - 9x^2 + 2x = 0 \implies x(4x^2 - 9x + 2) = 0$$ 3. Solve $x=0$ or quadratic $4x^2 - 9x + 2=0$: $$x = \frac{9 \pm \sqrt{81 - 32}}{8} = \frac{9 \pm \sqrt{49}}{8} = \frac{9 \pm 7}{8}$$ So, $$x=0, \quad x=\frac{9-7}{8} = \frac{2}{8} = 0.25, \quad x=\frac{9+7}{8} = 2$$ 4. Find second derivative: $$f''(x) = 12x^2 - 18x + 2$$ 5. Evaluate $f''$ at critical points: - At $x=0$: $f''(0) = 2 > 0$ (local minimum) - At $x=0.25$: $f''(0.25) = 12(0.0625) - 18(0.25) + 2 = 0.75 - 4.5 + 2 = -1.75 < 0$ (local maximum) - At $x=2$: $f''(2) = 12(4) - 18(2) + 2 = 48 - 36 + 2 = 14 > 0$ (local minimum) 6. Evaluate $f$ at critical points and endpoints: - $f(0) = -5$ - $f(0.25) = (0.25)^4 - 3(0.25)^3 + (0.25)^2 - 5 = 0.0039 - 0.0469 + 0.0625 - 5 = -4.9805$ - $f(2) = 16 - 24 + 4 - 5 = -9$ - $f(-5) = 625 + 375 + 25 - 5 = 1020$ - $f(5) = 625 - 375 + 25 - 5 = 270$ 7. Global max is $f(-5) = 1020$, global min is $f(2) = -9$. --- ### b) $f(x) = 4 - |x - 3|$ 1. The function is piecewise linear with a vertex at $x=3$. 2. For $x < 3$, $f(x) = 4 - (3 - x) = x + 1$ (increasing) 3. For $x > 3$, $f(x) = 4 - (x - 3) = 7 - x$ (decreasing) 4. At $x=3$, $f(3) = 4$ is a local maximum. 5. Evaluate at endpoints: - $f(-5) = 4 - | -5 - 3| = 4 - 8 = -4$ - $f(5) = 4 - |5 - 3| = 4 - 2 = 2$ 6. Global max is $4$ at $x=3$, global min is $-4$ at $x=-5$. --- ### c) $f(x) = e^{-x^2/2}$ 1. Derivative: $$f'(x) = e^{-x^2/2} \cdot (-x) = -x e^{-x^2/2}$$ 2. Set $f'(x)=0$: $$-x e^{-x^2/2} = 0 \implies x=0$$ 3. Second derivative: $$f''(x) = \frac{d}{dx}(-x e^{-x^2/2}) = -e^{-x^2/2} + x^2 e^{-x^2/2} = e^{-x^2/2}(x^2 - 1)$$ 4. Evaluate at $x=0$: $$f''(0) = e^0 (0 - 1) = -1 < 0$$ so local maximum at $x=0$. 5. Evaluate at endpoints: - $f(-5) = e^{-25/2} \approx 1.388 \times 10^{-6}$ - $f(5) = e^{-25/2} \approx 1.388 \times 10^{-6}$ 6. Global max is $f(0) = 1$, global min near zero at endpoints. --- ### d) $f(x) = \frac{x^2}{x-2}$ on $[-5,5]$ excluding $x=2$ 1. Derivative using quotient rule: $$f'(x) = \frac{2x(x-2) - x^2(1)}{(x-2)^2} = \frac{2x^2 - 4x - x^2}{(x-2)^2} = \frac{x^2 - 4x}{(x-2)^2}$$ 2. Set numerator zero: $$x^2 - 4x = x(x-4) = 0 \implies x=0, x=4$$ 3. Second derivative is complicated but test sign changes or use first derivative test: - For $x<0$, numerator positive (e.g., $x=-1$: $1 + 4 = 5 > 0$), denominator positive, so $f'(x)>0$ increasing. - Between $0$ and $4$, test $x=2.5$: numerator $2.5^2 - 4(2.5) = 6.25 - 10 = -3.75 < 0$, denominator positive, so $f'(x)<0$ decreasing. - For $x>4$, test $x=5$: numerator $25 - 20 = 5 > 0$, denominator positive, so $f'(x)>0$ increasing. 4. So local max at $x=0$, local min at $x=4$. 5. Evaluate function at critical points and endpoints: - $f(0) = 0$ - $f(4) = \frac{16}{2} = 8$ - $f(-5) = \frac{25}{-7} \approx -3.571$ - $f(5) = \frac{25}{3} \approx 8.333$ 6. Check behavior near 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 $f(x) \to +\infty$ 7. Global max is $f(5) \approx 8.333$, global min is unbounded below near $x=2^-$. --- **Final answers:** - a) Local minima at $x=0,2$, local max at $x=0.25$, global max $f(-5)=1020$, global min $f(2)=-9$. - b) Local and global max at $x=3$ with value 4, global min at $x=-5$ with value -4. - c) Local and global max at $x=0$ with value 1, global min near endpoints close to 0. - d) Local max at $x=0$ with value 0, local min at $x=4$ with value 8, global max at $x=5$ approx 8.333, no global min due to vertical asymptote.