1. **State the problem:** Determine which statement is not true about the graph of $$f(x) = x^{\frac{4}{3}} + 4x^{\frac{1}{3}} + 2$$. 2. **Recall the function and its derivatives:** $$f(x) = x^{\frac{4}{3}} + 4x^{\frac{1}{3}} + 2$$ First derivative: $$f'(x) = \frac{4}{3}x^{\frac{1}{3}} + \frac{4}{3}x^{-\frac{2}{3}} = \frac{4}{3} \left(x^{\frac{1}{3}} + x^{-\frac{2}{3}}\right)$$ Second derivative: $$f''(x) = \frac{4}{3} \left( \frac{1}{3}x^{-\frac{2}{3}} - \frac{2}{3}x^{-\frac{5}{3}} \right) = \frac{4}{9} \left(x^{-\frac{2}{3}} - 2x^{-\frac{5}{3}}\right)$$ 3. **Analyze concavity on $$(-\infty, 0)$$:** Rewrite $$f''(x)$$: $$f''(x) = \frac{4}{9} x^{-\frac{5}{3}} (x - 2)$$ For $$x < 0$$, note that $$x - 2 < 0$$ and $$x^{-\frac{5}{3}}$$ is negative because the exponent is negative and the base is negative (odd root). Multiplying two negatives gives positive, so: $$f''(x) > 0$$ on $$(-\infty, 0)$$, meaning $$f$$ is concave **upward** there. Therefore, the statement "$$f$$ is concave downward on $$(-\infty, 0)$$" is **not true**. 4. **Check for relative maxima and minima:** Set $$f'(x) = 0$$: $$\frac{4}{3} \left(x^{\frac{1}{3}} + x^{-\frac{2}{3}}\right) = 0 \implies x^{\frac{1}{3}} + x^{-\frac{2}{3}} = 0$$ Multiply both sides by $$x^{\frac{2}{3}}$$ (assuming $$x \neq 0$$): $$\cancel{x^{\frac{2}{3}}} \left(x^{\frac{1}{3}} + x^{-\frac{2}{3}}\right) = \cancel{x^{\frac{2}{3}}} \cdot 0 \implies x^{\frac{1}{3} + \frac{2}{3}} + 1 = 0$$ $$x^{1} + 1 = 0 \implies x = -1$$ 5. **Determine nature of critical point at $$x = -1$$:** Calculate $$f''(-1)$$: $$f''(-1) = \frac{4}{9} (-1)^{-\frac{5}{3}} (-1 - 2) = \frac{4}{9} (-1)^{-\frac{5}{3}} (-3)$$ Since $$(-1)^{-\frac{5}{3}} = -1$$ (odd root and negative exponent), $$f''(-1) = \frac{4}{9} \times (-1) \times (-3) = \frac{12}{9} = \frac{4}{3} > 0$$ Positive second derivative means a relative minimum at $$x = -1$$. 6. **Check continuity:** The function is composed of fractional powers with denominators 3, which are defined for all real $$x$$ (odd roots are defined for negative numbers), so $$f$$ is continuous on $$(-\infty, \infty)$$. 7. **Summary:** - $$f$$ is **not** concave downward on $$(-\infty, 0)$$ (it is concave upward). - $$f$$ has no relative maxima. - $$f$$ has a relative minimum at $$x = -1$$. - $$f$$ is continuous on $$(-\infty, \infty)$$. **Final answer:** The statement "$$f$$ is concave downward on $$(-\infty, 0)$$" is **not true**.