1. **Problem:** Find where the function $f(x) = x^3 - 1$ is increasing or decreasing using the Monotonicity Theorem. 2. **Formula and rules:** The Monotonicity Theorem states that if the derivative $f'(x) > 0$ on an interval, then $f$ is increasing there; if $f'(x) < 0$, then $f$ is decreasing. 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}(x^3 - 1) = 3x^2$$ 4. **Analyze the derivative:** Since $3x^2 \geq 0$ for all $x$, and equals zero only at $x=0$, the function is increasing everywhere except possibly at $x=0$. 5. **Conclusion:** - $f(x)$ is increasing on $(-\infty, 0)$ and $(0, \infty)$. - At $x=0$, the derivative is zero, so the function has a stationary point but does not decrease. --- 1. **Problem:** Find where the function $f(t) = t^3 + 3t^2 - 12$ is increasing or decreasing using the Monotonicity Theorem. 2. **Formula and rules:** Use the derivative test as above. 3. **Find the derivative:** $$f'(t) = \frac{d}{dt}(t^3 + 3t^2 - 12) = 3t^2 + 6t = 3t(t + 2)$$ 4. **Find critical points:** Set $f'(t) = 0$: $$3t(t + 2) = 0 \implies t=0 \text{ or } t=-2$$ 5. **Test intervals:** - For $t < -2$, pick $t=-3$: $f'(-3) = 3(-3)(-3+2) = 3(-3)(-1) = 9 > 0$ (increasing) - For $-2 < t < 0$, pick $t=-1$: $f'(-1) = 3(-1)(1) = -3 < 0$ (decreasing) - For $t > 0$, pick $t=1$: $f'(1) = 3(1)(3) = 9 > 0$ (increasing) 6. **Conclusion:** - Increasing on $(-\infty, -2)$ and $(0, \infty)$ - Decreasing on $(-2, 0)$ --- **Final answers:** - For $f(x) = x^3 - 1$: increasing on $(-\infty, 0) \cup (0, \infty)$ - For $f(t) = t^3 + 3t^2 - 12$: increasing on $(-\infty, -2) \cup (0, \infty)$, decreasing on $(-2, 0)$