1. **Stating the problem:** We have a function $f : \mathbb{R} \to \mathbb{R}$ defined by $$f(x) = \sqrt[3]{x^2}(3x - 7) = x^{\frac{2}{3}}(3x - 7).$$ We want to find values $a$ and $b$ such that $f$ is increasing on $(-\infty, a)$ and $(b, \infty)$, and then compute $7a + 15b$. 2. **Formula and rules:** To determine where $f$ is increasing, we find its derivative $f'(x)$ and analyze its sign. The function is increasing where $f'(x) > 0$. 3. **Find the derivative:** Using the product rule: $$f'(x) = \frac{d}{dx}\left(x^{\frac{2}{3}}\right)(3x - 7) + x^{\frac{2}{3}} \cdot \frac{d}{dx}(3x - 7).$$ Calculate each derivative: $$\frac{d}{dx}x^{\frac{2}{3}} = \frac{2}{3}x^{-\frac{1}{3}},$$ $$\frac{d}{dx}(3x - 7) = 3.$$ So, $$f'(x) = \frac{2}{3}x^{-\frac{1}{3}}(3x - 7) + 3x^{\frac{2}{3}} = \frac{2}{3}x^{-\frac{1}{3}}(3x - 7) + 3x^{\frac{2}{3}}.$$ 4. **Simplify the derivative:** Multiply terms: $$f'(x) = \frac{2}{3}x^{-\frac{1}{3}} \cdot 3x - \frac{2}{3}x^{-\frac{1}{3}} \cdot 7 + 3x^{\frac{2}{3}} = 2x^{\frac{2}{3}} - \frac{14}{3}x^{-\frac{1}{3}} + 3x^{\frac{2}{3}}.$$ Combine like terms: $$f'(x) = (2 + 3)x^{\frac{2}{3}} - \frac{14}{3}x^{-\frac{1}{3}} = 5x^{\frac{2}{3}} - \frac{14}{3}x^{-\frac{1}{3}}.$$ Rewrite with common denominator: $$f'(x) = \frac{15x - 14}{3x^{\frac{1}{3}}}.$$ 5. **Analyze the sign of $f'(x)$:** The denominator $3x^{\frac{1}{3}}$ has the same sign as $x^{\frac{1}{3}}$, which is positive if $x > 0$, negative if $x < 0$, and zero at $x=0$. The numerator is $15x - 14$. 6. **Find critical points:** Set numerator and denominator to zero: - Numerator zero: $15x - 14 = 0 \Rightarrow x = \frac{14}{15} \approx 0.9333$ - Denominator zero: $x = 0$ 7. **Sign chart:** - For $x < 0$, denominator is negative. - For $0 < x < \frac{14}{15}$, numerator is negative. - For $x > \frac{14}{15}$, numerator is positive. Evaluate $f'(x)$ sign in intervals: - $(-\infty, 0)$: numerator $15x - 14$ is negative (since $x$ negative), denominator negative, so $f'(x) = \frac{negative}{negative} = positive$. - $(0, \frac{14}{15})$: numerator negative, denominator positive, so $f'(x)$ negative. - $(\frac{14}{15}, \infty)$: numerator positive, denominator positive, so $f'(x)$ positive. 8. **Conclusion:** $f$ is increasing on $(-\infty, 0)$ and $(\frac{14}{15}, \infty)$, so $a = 0$ and $b = \frac{14}{15}$. 9. **Calculate $7a + 15b$:** $$7a + 15b = 7 \cdot 0 + 15 \cdot \frac{14}{15} = 14.$$ **Final answer:** 14 (Option B).