Let's solve the problem **step-by-step**. ### 1. **Given Function** \[ f(x) = \frac{x^4}{2} + 0.5x^2 + 42x^2 \] Combine like terms: \[ f(x) = \frac{1}{2}x^4 + 42.5x^2 \] --- ### 2. **Find the Second Derivative** First, find the first derivative: \[ f'(x) = 2x^3 + 2 \times 42.5x = 2x^3 + 85x \] Now, the second derivative: \[ f''(x) = 6x^2 + 85 \] --- ### 3. **Find Inflection Points** Inflection points occur where \( f''(x) = 0 \): \[ 6x^2 + 85 = 0 \] \[ 6x^2 = -85 \] \[ x^2 = -\frac{85}{6} \] But this is **impossible** for real numbers since the right side is negative. This suggests a mistake in simplification or transcription. Let's double-check the original function from the image: From the image: \[ f(x) = \frac{x^4}{2} + 0.5x^3 + 42x^2 \] So let's use this: \[ f(x) = \frac{1}{2}x^4 + 0.5x^3 + 42x^2 \] Find the first derivative: \[ f'(x) = 2x^3 + 1.5x^2 + 84x \] Second derivative: \[ f''(x) = 6x^2 + 3x + 84 \] Set \( f''(x) = 0 \): \[ 6x^2 + 3x + 84 = 0 \] Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 6 \), \( b = 3 \), \( c = 84 \): \[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 6 \cdot 84}}{2 \cdot 6} \] \[ x = \frac{-3 \pm \sqrt{9 - 2016}}{12} \] \[ x = \frac{-3 \pm \sqrt{-2007}}{12} \] Again, the discriminant is negative. **There are no real inflection points**. --- ### 4. **Verify with the Graph** Looking at the graph, there appear to be two inflection points (where the concavity changes). Let's try a more careful derivative for \( f(x) = \frac{x^4}{2} + 0.5x^2 + 42x^2 \): But the second term is \( 0.5x^2 \), not \( 0.5x^3 \). So, \( f(x) = \frac{x^4}{2} + 0.5x^2 + 42x^2 = \frac{x^4}{2} + 42.5x^2 \). First derivative: \[ f'(x) = 2x^3 + 85x \] Second derivative: \[ f''(x) = 6x^2 + 85 \] Set \( f''(x) = 0 \): \[ 6x^2 + 85 = 0 \implies x^2 = -\frac{85}{6} \] Again, no real solution. But the question and graph suggest otherwise. ### 5. **Possible Typo in the Function?** If the middle term is \( 0.5x^3 \): \[ f(x) = \frac{x^4}{2} + 0.5x^3 + 42x^2 \] First derivative: \[ f'(x) = 2x^3 + 1.5x^2 + 84x \] Second derivative: \[ f''(x) = 6x^2 + 3x + 84 \] Set \( f''(x) = 0 \): \[ 6x^2 + 3x + 84 = 0 \] \[ x = \frac{-3 \pm \sqrt{9 - 2016}}{12} \] \[ x = \frac{-3 \pm \sqrt{-2007}}{12} \] No real solutions. ### 6. **Conclusion** **Based on the function as written, the second derivative does not have real roots, so there are no real inflection points.** However, **the graph clearly shows two inflection points**. There must be a typo in the function provided. If you can clarify the function, I can solve it step by step. --- #### **Based on Graph (Visual Guess):** - **Interval \((-\infty, C)\):** Concave down (the graph is curving downward). - **Interval \((C, D)\):** Concave up (the graph is curving upward). - **Interval \((D, \infty)\):** Concave down (the graph is curving downward). **If you want the precise inflection points and math, please check the function for typos.** --- ### **Summary Table (Based on Graph)** | Interval | Concavity | |----------------|-------------| | \((-\infty, C)\) | Concave Down | | \((C, D)\) | Concave Up | | \((D, \infty)\) | Concave Down | **If you clarify the function, I can give you the exact values for C and D!**

Let's carefully revisit the problem with the correct function: Given function: \[ f(x) = \frac{x^4}{2} + 0.5x^2 + 42x^2 \] **Step 1: Simplify the function** Combine like terms: \[ f(x) = \frac{x^4}{2} + (0.5x^2 + 42x^2) = \frac{x^4}{2} + 42.5x^2 \] --- **Step 2: Find the first derivative \(f'(x)\)** Differentiate term-by-term: \[ f'(x) = \frac{d}{dx} \left( \frac{x^4}{2} \right) + \frac{d}{dx} (42.5x^2) \] \[ f'(x) = 2x^3 + 2 \times 42.5 x = 2x^3 + 85x \] --- **Step 3: Find the second derivative \(f''(x)\)** Differentiate \(f'(x)\): \[ f''(x) = \frac{d}{dx} (2x^3 + 85x) = 6x^2 + 85 \] --- **Step 4: Find inflection points** Inflection points occur where \(f''(x) = 0\): \[ 6x^2 + 85 = 0 \] \[ 6x^2 = -85 \] \[ x^2 = -\frac{85}{6} \] Since \(x^2\) cannot be negative for real \(x\), **there are no real solutions** to \(f''(x)=0\). **Conclusion:** - The second derivative \(f''(x) = 6x^2 + 85\) is always positive because \(6x^2 \geq 0\) and \(85 > 0\). - Therefore, the function is **concave up everywhere**, and **there are no inflection points**. --- ### **Final answer:** **The function has no points of inflection because the second derivative is always positive.** **Thus, the graph does not change concavity, and the second derivative is never zero for real \(x\).** --- **Note:** The initial confusion was due to misreading the function. The correct simplified form is: \[ f(x) = \frac{x^4}{2} + 42.5x^2 \] with second derivative: \[ f''(x) = 6x^2 + 85 > 0 \quad \text{for all real } x \] **Hence, no inflection points exist.**

Let's carefully analyze the given function: \[ f(x) = \frac{x^4}{2} + 0.5x^3 + 42x^2 \] --- ### Step 1: Find the first derivative \(f'(x)\) Differentiate term-by-term: \[ f'(x) = \frac{d}{dx} \left(\frac{x^4}{2}\right) + \frac{d}{dx}(0.5x^3) + \frac{d}{dx}(42x^2) \] \[ f'(x) = 2x^3 + 1.5x^2 + 84x \] --- ### Step 2: Find the second derivative \(f''(x)\) Differentiate \(f'(x)\): \[ f''(x) = \frac{d}{dx}(2x^3 + 1.5x^2 + 84x) = 6x^2 + 3x + 84 \] --- ### Step 3: Find inflection points (where \(f''(x) = 0\)) Set the second derivative to zero: \[ 6x^2 + 3x + 84 = 0 \] Divide entire equation by 3 to simplify: \[ 2x^2 + x + 28 = 0 \] Apply quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a=2\), \(b=1\), \(c=28\): \[ x = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times 28}}{2 \times 2} \] \[ x = \frac{-1 \pm \sqrt{1 - 224}}{4} \] \[ x = \frac{-1 \pm \sqrt{-223}}{4} \] Since the discriminant \(\sqrt{-223}\) is imaginary, **there are no real solutions**. --- ### **Conclusion:** - The second derivative \(f''(x) = 6x^2 + 3x + 84\) is always positive because the quadratic has no real roots and opens upward (coefficient of \(x^2\) is positive). - Therefore, **the function is concave up everywhere**, and **there are no inflection points**. --- ### **Final answer:** **The function \(f(x) = \frac{x^4}{2} + 0.5x^3 + 42x^2\) has no points of inflection because its second derivative never equals zero for real \(x\).** **Graphically, the function is always concave up, and the concavity does not change.**

Let's carefully analyze the function and find the inflection points step-by-step. --- ### Given Function: \[ f(x) = \frac{x^4}{2} + 0.5x^3 + 42x^2 \] --- ### **Step 1: Find the First Derivative \(f'(x)\)** Differentiate each term: - \(\frac{x^4}{2}\): \[ \frac{d}{dx} \left( \frac{x^4}{2} \right) = \frac{4x^3}{2} = 2x^3 \] - \(0.5x^3\): \[ \frac{d}{dx} (0.5x^3) = 0.5 \times 3x^2 = 1.5x^2 \] - \(42x^2\): \[ \frac{d}{dx} (42x^2) = 2 \times 42 x = 84x \] **Therefore:** \[ \boxed{ f'(x) = 2x^3 + 1.5x^2 + 84x } \] --- ### **Step 2: Find the Second Derivative \(f''(x)\)** Differentiate \(f'(x)\): - \(2x^3\): \[ \frac{d}{dx} (2x^3) = 6x^2 \] - \(1.5x^2\): \[ \frac{d}{dx} (1.5x^2) = 3x \] - \(84x\): \[ \frac{d}{dx} (84x) = 84 \] **Thus:** \[ \boxed{ f''(x) = 6x^2 + 3x + 84 } \] --- ### **Step 3: Find the Inflection Points (Set \(f''(x) = 0\))** Solve: \[ 6x^2 + 3x + 84 = 0 \] Divide through by 3 to simplify: \[ 2x^2 + x + 28 = 0 \] Apply quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a=2\), \(b=1\), \(c=28\): \[ x = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times 28}}{2 \times 2} \] Calculate discriminant: \[ \Delta = 1 - 4 \times 2 \times 28 = 1 - 224 = -223 \] Since \(\Delta < 0\), **there are no real solutions**. --- ### **Step 4: Interpret the Results** Because the second derivative \(f''(x) = 6x^2 + 3x + 84\) is a quadratic with a positive leading coefficient and **discriminant negative**, it **never crosses zero**. Let's check the sign of \(f''(x)\): - For all real \(x\), \(6x^2 \geq 0\), and adding 84 makes the entire expression **always positive**. **Therefore:** \[ f''(x) > 0 \quad \text{for all real } x \] This implies: - The function is **concave up everywhere**. - **No points of inflection** exist where the concavity changes. --- ### **Final Answer:** \[ \boxed{ \text{The function } f(x) = \frac{x^4}{2} + 0.5x^3 + 42x^2 \text{ has no inflection points because } f''(x) > 0 \text{ for all } x. } \] **In conclusion:** - The graph is always concave upward. - The second derivative does not equal zero at any real point. - Therefore, **no points where the concavity changes**. --- **If you need further analysis or clarification, feel free to ask!**