Question

The maximum value of the function \( f(x) = 3x^2 - 2x^3 \) for \( x>0 \) is _________ .

Hint:

To find the maximum or minimum value of a function, take the derivative, set it equal to 0, and use the second derivative test to determine the nature of the critical points.

Answer 1

Correct Answer: 1

To find the maximum value of the function, we first find its critical points by taking the derivative of \( f(x) \) and setting it equal to 0: \[ f'(x) = 6x - 6x^2. \] Set \( f'(x) = 0 \): \[ 6x - 6x^2 = 0 \quad \Rightarrow \quad 6x(1 - x) = 0. \] So, the critical points are \( x = 0 \) and \( x = 1 \). Since we are interested in \( x>0 \), we have \( x = 1 \). Now, we check the second derivative to determine if \( x = 1 \) is a maximum: \[ f''(x) = 6 - 12x. \] At \( x = 1 \): \[ f''(1) = 6 - 12(1) = -6. \] Since \( f''(1)<0 \), \( x = 1 \) is a local maximum. Finally, evaluate \( f(x) \) at \( x = 1 \): \[ f(1) = 3(1)^2 - 2(1)^3 = 3 - 2 = 1. \] Thus, the maximum value of the function is 1.