To determine the maximum value of the function \( y = 3x^2 - x^3 \), we will follow a systematic approach involving calculus. Here’s how we can find the maximum value step-by-step: ### Step 1: Find the derivative To find the maximum value of the function, we first need to find its derivative. The derivative of a function gives us the slope of the tangent line at any point on the curve. We can find the derivative of \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx}(3x^2 - x^3) \] Using the power rule, we differentiate each term: \[ \frac{dy}{dx} = 6x - 3x^2 \] ### Step 2: Set the derivative to zero To find the critical points (where the function may have a maximum or minimum), we set the derivative equal to zero: \[ 6x - 3x^2 = 0 \] Factoring out the common term: \[ 3x(2 - x) = 0 \] This gives us two solutions: 1. \( 3x = 0 \) → \( x = 0 \) 2. \( 2 - x = 0 \) → \( x = 2 \) ### Step 3: Determine the nature of the critical points Next, we need to determine whether these critical points correspond to a maximum or minimum. We can do this by using the second derivative test. First, we find the second derivative: \[ \frac{d^2y}{dx^2} = \frac{d}{dx}(6x - 3x^2) = 6 - 6x \] Now we evaluate the second derivative at our critical points: 1. For \( x = 0 \): \[ \frac{d^2y}{dx^2} = 6 - 6(0) = 6 \quad (\text{positive, indicating a local minimum}) \] 2. For \( x = 2 \): \[ \frac{d^2y}{dx^2} = 6 - 6(2) = 6 - 12 = -6 \quad (\text{negative, indicating a local maximum}) \] ### Step 4: Calculate the maximum value Since \( x = 2 \) is a local maximum, we can find the maximum value of \( y \) by substituting \( x = 2 \) back into the original function: \[ y = 3(2^2) - (2^3) = 3(4) - 8 = 12 - 8 = 4 \] ### Conclusion The maximum value of \( y = 3x^2 - x^3 \) occurs at \( x = 2 \) and is equal to 4. ### Explanation of Options - **Option A: Zero** - This is incorrect because we found a maximum value of 4, not zero. - **Option B: 2** - This is incorrect as the maximum value is greater than 2. - **Option C: 4** - This is correct as we calculated the maximum value to be 4. - **Option D: 6** - This is incorrect because the maximum value we found is less than 6. ### Revision Summary - To find the maximum value of a function, first find its derivative and set it to zero to locate critical points. - Use the second derivative test to determine whether each critical point is a maximum or minimum. - Substitute the critical points back into the original function to find the maximum value. - In this case, the maximum value of \( y = 3x^2 - x^3 \) is 4 at \( x = 2 \).