Let's solve the problem step-by-step: ### **Step 1: Find the second derivative** Given: \[ f(x) = x^3 + 3x^2 \] First derivative: \[ f'(x) = 3x^2 + 6x \] Second derivative: \[ f''(x) = 6x + 6 \] --- ### **Step 2: Set the second derivative greater than zero (for concave up)** A function is **concave up** where \( f''(x) > 0 \): \[ 6x + 6 > 0 \] \[ 6x > -6 \] \[ x > -1 \] --- ### **Step 3: Write the answer in interval notation** The interval where the function is concave up is: \[ \boxed{(-1, \infty)} \] --- ### **Final Answer** \[ \boxed{(-1, \infty)} \] **Explanation:** The function \( f(x) = x^3 + 3x^2 \) is concave up for all \( x > -1 \).

Let's carefully analyze the problem step-by-step, including all detailed calculations and explanations. --- ### **Given function:** \[ f(z) = z^3 + 3z^2 \] *(Note: The original problem appears to involve \(f(z)\), but the calculations are similar to those with \(f(x)\).)* --- ### **Step 1: Find the first derivative \(f'(z)\)** Differentiate \(f(z) = z^3 + 3z^2\): \[ f'(z) = \frac{d}{dz}(z^3) + \frac{d}{dz}(3z^2) = 3z^2 + 6z \] --- ### **Step 2: Find the second derivative \(f''(z)\)** Differentiate \(f'(z)\): \[ f''(z) = \frac{d}{dz}(3z^2 + 6z) = 6z + 6 \] --- ### **Step 3: Determine where \(f(z)\) is concave up** A function is **concave up** where its second derivative is **positive**: \[ f''(z) > 0 \] Substitute \(f''(z)\): \[ 6z + 6 > 0 \] Solve for \(z\): \[ 6z > -6 \] \[ z > -1 \] --- ### **Step 4: Express the interval in interval notation** The function is concave up for all \(z > -1\). Therefore, the interval is: \[ \boxed{(-1, \infty)} \] --- ### **Final summarized answer:** \[ \boxed{(-1, \infty)} \] --- ### **Additional notes:** - The critical point at \(z = -1\) is where the concavity changes. - For \(z < -1\), \(f''(z) < 0\), so the function is concave down. - For \(z > -1\), \(f''(z) > 0\), so the function is concave up. **This completes the step-by-step detailed solution!**