Question

The maximum value of the function \( h(x) = -x^3 + 2x^2 \) in the interval \([-1, 1.5]\) is equal to ......... (rounded off to 1 decimal place).

Hint:

To find the maximum or minimum of a function on a closed interval, check the critical points and the endpoints of the interval.

Answer 1

Correct Answer: 2.9

We are given the function \( h(x) = -x^3 + 2x^2 \). First, find the first derivative of the function: \[ h'(x) = -3x^2 + 4x. \] Now, find the critical points by setting \( h'(x) = 0 \): \[ -3x^2 + 4x = 0 \quad \Rightarrow \quad x(4 - 3x) = 0. \] This gives \( x = 0 \) and \( x = \frac{4}{3} \). These are the critical points. Next, calculate the second derivative: \[ h''(x) = -6x + 4. \] Check the concavity at each critical point: At \( x = 0 \), \( h''(0) = 4 \), which means \( x = 0 \) is a point of minima. At \( x = \frac{4}{3} \), \( h''\left(\frac{4}{3}\right) = -\frac{4}{3} \), which means \( x = \frac{4}{3} \) is a point of maxima. Now, check the value of \( h(x) \) at the critical points and at the boundaries of the interval: - At \( x = -1 \), \( h(-1) = 3 \). - At \( x = 0 \), \( h(0) = 0 \). - At \( x = \frac{4}{3} \), \( h\left(\frac{4}{3}\right) = 1.125 \). - At \( x = 1.5 \), \( h(1.5) = -1.125 \). Thus, the maximum value of \( h(x) \) is 3, which occurs at \( x = -1 \).