1. **Problem 1:** Analyze and plot the graph of $$f(x) = x^{\frac{5}{3}} - 4x^{\frac{3}{5}}$$ for all real $$x$$. 2. **Problem 2:** Analyze and plot the graph of $$f(x) = x + \frac{3}{x}$$ for $$x \neq 0$$. --- ### Problem 1: $$f(x) = x^{\frac{5}{3}} - 4x^{\frac{3}{5}}$$ 1. **State the problem:** We want to analyze the function including domain, intercepts, critical points, and behavior. 2. **Domain:** Since fractional exponents with odd denominators allow negative bases, domain is all real numbers $$\mathbb{R}$$. 3. **Find intercepts:** - At $$x=0$$, $$f(0) = 0 - 0 = 0$$, so the graph passes through the origin. 4. **Find derivative:** $$f'(x) = \frac{5}{3}x^{\frac{5}{3} - 1} - 4 \cdot \frac{3}{5} x^{\frac{3}{5} - 1} = \frac{5}{3}x^{\frac{2}{3}} - \frac{12}{5}x^{-\frac{2}{5}}$$ 5. **Critical points:** Set $$f'(x) = 0$$: $$\frac{5}{3}x^{\frac{2}{3}} = \frac{12}{5}x^{-\frac{2}{5}}$$ Multiply both sides by $$x^{\frac{2}{5}}$$: $$\frac{5}{3}x^{\frac{2}{3} + \frac{2}{5}} = \frac{12}{5}$$ Calculate exponent sum: $$\frac{2}{3} + \frac{2}{5} = \frac{10}{15} + \frac{6}{15} = \frac{16}{15}$$ So: $$\frac{5}{3}x^{\frac{16}{15}} = \frac{12}{5}$$ Solve for $$x^{\frac{16}{15}}$$: $$x^{\frac{16}{15}} = \frac{12}{5} \cdot \frac{3}{5} = \frac{36}{25}$$ 6. **Solve for $$x$$:** $$x = \left(\frac{36}{25}\right)^{\frac{15}{16}}$$ 7. **Second derivative test:** $$f''(x) = \frac{5}{3} \cdot \frac{2}{3} x^{\frac{2}{3} - 1} + \frac{12}{5} \cdot \frac{2}{5} x^{-\frac{2}{5} - 1} = \frac{10}{9} x^{-\frac{1}{3}} + \frac{24}{25} x^{-\frac{7}{5}}$$ Evaluate sign at critical points to determine maxima or minima. --- ### Problem 2: $$f(x) = x + \frac{3}{x}$$, $$x \neq 0$$ 1. **Domain:** All real numbers except $$x=0$$. 2. **Intercepts:** - $$y$$-intercept does not exist (undefined at $$x=0$$). - $$x$$-intercepts: Solve $$x + \frac{3}{x} = 0$$ Multiply both sides by $$x$$: $$x^2 + 3 = 0$$ No real roots since $$x^2 = -3$$ is impossible. 3. **Derivative:** $$f'(x) = 1 - \frac{3}{x^2}$$ 4. **Critical points:** Set $$f'(x) = 0$$: $$1 = \frac{3}{x^2} \Rightarrow x^2 = 3 \Rightarrow x = \pm \sqrt{3}$$ 5. **Second derivative:** $$f''(x) = \frac{6}{x^3}$$ 6. **Classify critical points:** - At $$x = \sqrt{3}$$, $$f''(\sqrt{3}) = \frac{6}{(\sqrt{3})^3} > 0$$, so local minimum. - At $$x = -\sqrt{3}$$, $$f''(-\sqrt{3}) < 0$$, so local maximum. --- ### Summary: - Problem 1 domain: $$\mathbb{R}$$, critical point at $$x = \left(\frac{36}{25}\right)^{\frac{15}{16}}$$. - Problem 2 domain: $$\mathbb{R} \setminus \{0\}$$, local max at $$x = -\sqrt{3}$$, local min at $$x = \sqrt{3}$$.