1. **Problem statement:** Find the values of $x$ for which the derived function $f'(x) = 3x^2$ is zero, and prepare a gradient table to determine whether these points are maxima, minima, or points of inflexion. 2. **Formula and rules:** To find critical points, solve $f'(x) = 0$. Then analyze the sign of $f'(x)$ around these points to classify them: - If $f'(x)$ changes from positive to negative, it's a maximum. - If $f'(x)$ changes from negative to positive, it's a minimum. - If $f'(x)$ does not change sign, it's a point of inflexion. 3. **Solve for zeros:** $$3x^2 = 0$$ Divide both sides by 3: $$x^2 = 0$$ Take square root: $$x = 0$$ 4. **Gradient table:** - For $x < 0$, pick $x = -1$: $f'(-1) = 3(-1)^2 = 3 > 0$ - For $x > 0$, pick $x = 1$: $f'(1) = 3(1)^2 = 3 > 0$ | Interval | Sign of $f'(x)$ | Behavior | |---------|-----------------|----------| | $(-\infty, 0)$ | Positive | Increasing | | $0$ | Zero | Critical point | | $(0, \infty)$ | Positive | Increasing | Since $f'(x)$ is positive on both sides of $x=0$, the function is increasing before and after $0$, so $x=0$ is a point of inflexion. 5. **Final answer:** - The derived function $f'(x) = 3x^2$ is zero at $x=0$. - At $x=0$, the function has a point of inflexion, not a maximum or minimum.