1. **Problem:** Determine the intervals where the function $f(x) = x^3 - 3x^2 + 4$ is concave up or concave down. 2. **Formula and rules:** - The concavity of a function is determined by the sign of its second derivative $f''(x)$. - If $f''(x) > 0$, the function is concave up on that interval. - If $f''(x) < 0$, the function is concave down on that interval. 3. **Find the first derivative:** $$f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 4) = 3x^2 - 6x$$ 4. **Find the second derivative:** $$f''(x) = \frac{d}{dx}(3x^2 - 6x) = 6x - 6$$ 5. **Find critical points for concavity by setting $f''(x) = 0$:** $$6x - 6 = 0$$ $$6x = 6$$ $$x = 1$$ 6. **Test intervals around $x=1$ to determine concavity:** - For $x < 1$, choose $x=0$: $f''(0) = 6(0) - 6 = -6 < 0$ so concave down. - For $x > 1$, choose $x=2$: $f''(2) = 6(2) - 6 = 6 > 0$ so concave up. 7. **Answer:** - The function is concave down on $(-\infty, 1)$. - The function is concave up on $(1, \infty)$. **Multiple Choice Question:** Where is the function $f(x) = x^3 - 3x^2 + 4$ concave up? A) $(-\infty, 1)$ B) $(1, \infty)$ C) $(-\infty, 0)$ D) $(0, 1)$