1. **Problem Statement:** Determine the intervals where the function $f(x) = x^3 - 6x^2 + 15$ is increasing or decreasing using the first derivative test. 2. **Formula and Rules:** To find where a function is increasing or decreasing, we use the first derivative $f'(x)$. The function is increasing where $f'(x) > 0$ and decreasing where $f'(x) < 0$. Critical points occur where $f'(x) = 0$ or is undefined. 3. **Find the first derivative:** $$f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 15) = 3x^2 - 12x$$ 4. **Find critical points by solving $f'(x) = 0$:** $$3x^2 - 12x = 0$$ $$3x(x - 4) = 0$$ So, $x = 0$ or $x = 4$. 5. **Test intervals around critical points:** - For $x < 0$, pick $x = -1$: $f'(-1) = 3(-1)^2 - 12(-1) = 3 + 12 = 15 > 0$ (increasing) - For $0 < x < 4$, pick $x = 2$: $f'(2) = 3(2)^2 - 12(2) = 12 - 24 = -12 < 0$ (decreasing) - For $x > 4$, pick $x = 5$: $f'(5) = 3(5)^2 - 12(5) = 75 - 60 = 15 > 0$ (increasing) 6. **Conclusion:** - The function is increasing on $(-\infty, 0)$ and $(4, \infty)$. - The function is decreasing on $(0, 4)$. 7. **First derivative test:** - At $x=0$, $f'(x)$ changes from positive to negative, so $f$ has a local maximum at $x=0$. - At $x=4$, $f'(x)$ changes from negative to positive, so $f$ has a local minimum at $x=4$. **Final answer:** - Increasing on $(-\infty, 0) \cup (4, \infty)$ - Decreasing on $(0, 4)$ - Local maximum at $x=0$ - Local minimum at $x=4$