1. **Problem 7:** Identify the graph representing the function $$f(x) = x(x - 1)(x - 3)$$. 2. **Step 1:** Expand the function to understand its shape. $$f(x) = x(x - 1)(x - 3) = x[(x - 1)(x - 3)] = x[x^2 - 4x + 3] = x^3 - 4x^2 + 3x$$ 3. **Step 2:** Analyze the roots and behavior. - Roots are at $$x=0, 1, 3$$. - Since the leading term is $$x^3$$, the graph behaves like a cubic with positive leading coefficient: it goes to $$- ext{infinity}$$ as $$x o - ext{infinity}$$ and $$+ ext{infinity}$$ as $$x o + ext{infinity}$$. 4. **Step 3:** Check the graph descriptions. - Graph a: starts below zero near $$x=-3$$, crosses x-axis at 0, local max near 1, local min near 2, rises sharply near 3. - This matches the roots and cubic behavior. **Answer for 7:** Graph a. --- 5. **Problem 8:** Find critical points of $$f(x) = 12x - x^3$$. 6. **Step 1:** Find derivative: $$f'(x) = 12 - 3x^2$$ 7. **Step 2:** Set derivative to zero to find critical points: $$12 - 3x^2 = 0 \implies 3x^2 = 12 \implies x^2 = 4 \implies x = \pm 2$$ 8. **Step 3:** Find corresponding $$y$$ values: $$f(2) = 12(2) - 2^3 = 24 - 8 = 16$$ $$f(-2) = 12(-2) - (-2)^3 = -24 + 8 = -16$$ **Answer for 8:** (2, 16) and (-2, -16) (option a). --- 9. **Problem 9:** Find critical point of $$f(x) = x(x - 6)$$. 10. **Step 1:** Expand: $$f(x) = x^2 - 6x$$ 11. **Step 2:** Derivative: $$f'(x) = 2x - 6$$ 12. **Step 3:** Set derivative to zero: $$2x - 6 = 0 \implies x = 3$$ 13. **Step 4:** Find $$y$$ value: $$f(3) = 3(3 - 6) = 3(-3) = -9$$ **Answer for 9:** (3, -9) (option c). --- 14. **Problem 10:** Find $$f''(x)$$ if $$f(x) = \sqrt[n]{x} = x^{1/n}$$. 15. **Step 1:** First derivative: $$f'(x) = \frac{1}{n} x^{\frac{1}{n} - 1}$$ 16. **Step 2:** Second derivative: $$f''(x) = \frac{1}{n} \left(\frac{1}{n} - 1\right) x^{\frac{1}{n} - 2} = \frac{1 - n}{n^2} x^{\frac{1}{n} - 2}$$ 17. **Step 3:** Rewrite using root notation: $$x^{\frac{1}{n} - 2} = x^{\frac{1 - 2n}{n}} = \sqrt[n]{x^{1 - 2n}}$$ 18. **Answer for 10:** $$\frac{1 - n}{n^2} \sqrt[n]{x^{1 - 2n}}$$ (option b).