Question

Find the maximum value of the function \( f(x) = -x^3 + 2x^2 \) in the interval \( [-1, 1.5] \).

Hint:

To find the maximum or minimum values of a function on a given interval, always evaluate the function at critical points and endpoints, then compare the values.

Answer 1

The function given is: \[ f(x) = -x^3 + 2x^2 \] To find the maximum value of \( f(x) \) in the interval \( [-1, 1.5] \), we first find the derivative of the function: \[ f'(x) = \frac{d}{dx}(-x^3 + 2x^2) = -3x^2 + 4x \] Next, we set the derivative equal to zero to find the critical points: \[ -3x^2 + 4x = 0 \] Factoring: \[ x(-3x + 4) = 0 \] This gives two solutions: \[ x = 0 \quad \text{or} \quad x = \frac{4}{3} \] Now, we evaluate the function at the critical points and at the endpoints of the interval \( [-1, 1.5] \): 1. At \( x = -1 \): \[ f(-1) = -(-1)^3 + 2(-1)^2 = 1 + 2 = 3 \] 2. At \( x = 0 \): \[ f(0) = -(0)^3 + 2(0)^2 = 0 \] 3. At \( x = \frac{4}{3} \): \[ f\left(\frac{4}{3}\right) = -\left(\frac{4}{3}\right)^3 + 2\left(\frac{4}{3}\right)^2 = -\frac{64}{27} + \frac{32}{9} = -\frac{64}{27} + \frac{96}{27} = \frac{32}{27} \approx 1.185 \] 4. At \( x = 1.5 \): \[ f(1.5) = -(1.5)^3 + 2(1.5)^2 = -3.375 + 4.5 = 1.125 \] Now, comparing these values: \[ f(-1) = 3, \quad f(0) = 0, \quad f\left(\frac{4}{3}\right) \approx 1.185, \quad f(1.5) = 1.125 \] The maximum value occurs at \( x = -1 \), and the maximum value is \( \mathbf{3} \).