1. **Problem Statement:** Determine if the function $f(x) = x^3$ has a turning point. 2. **Formula and Rules:** A turning point occurs where the first derivative $f'(x)$ changes sign, typically at critical points where $f'(x) = 0$. 3. **Find the first derivative:** $$f'(x) = \frac{d}{dx} x^3 = 3x^2$$ 4. **Find critical points by setting $f'(x) = 0$:** $$3x^2 = 0 \implies x = 0$$ 5. **Check the second derivative to determine the nature of the critical point:** $$f''(x) = \frac{d}{dx} 3x^2 = 6x$$ 6. **Evaluate $f''(0)$:** $$f''(0) = 6 \times 0 = 0$$ 7. Since $f''(0) = 0$, the second derivative test is inconclusive. We analyze the behavior of $f'(x)$ around $x=0$: - For $x < 0$, $f'(x) = 3x^2 > 0$ - For $x > 0$, $f'(x) = 3x^2 > 0$ 8. The derivative does not change sign around $x=0$, so $x=0$ is not a turning point. **Final answer:** The function $f(x) = x^3$ does not have a turning point.