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).

• A. \( 3 \)
• B. \( 2.5 \)
• C. \( 2.125 \)
• D. \( 1.125 \)

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

The Correct Option is A

Given Function:

\[ h(x) = -x^3 + 2x^2 \]

Step 1: Find the First Derivative

Differentiate \( h(x) \):

\[ h'(x) = -3x^2 + 4x \]

Step 2: Find Critical Points

Set \( h'(x) = 0 \) to find critical points:

\[ -3x^2 + 4x = 0 \]

Factoring:

\[ x(4 - 3x) = 0 \]

So, the critical points are:

\[ x = 0, \quad x = \frac{4}{3} \]

Step 3: Second Derivative Test

Compute the second derivative:

\[ h''(x) = -6x + 4 \]

Evaluate at \( x = 0 \):

\[ h''(0) = -6(0) + 4 = 4 \quad (\text{Positive} \Rightarrow \text{Local Minimum}) \]

Evaluate at \( x = \frac{4}{3} \):

\[ h''\left(\frac{4}{3}\right) = -6\left(\frac{4}{3}\right) + 4 = -4 \quad (\text{Negative} \Rightarrow \text{Local Maximum}) \]

Step 4: Compute Function Values

Evaluate \( h(x) \) at critical points and boundary points:

• \( h(-1) = -(-1)^3 + 2(-1)^2 = 1 + 2 = 3 \)
• \( h(0) = -(0)^3 + 2(0)^2 = 0 \)
• \( h\left(\frac{4}{3}\right) = -\left(\frac{4}{3}\right)^3 + 2\left(\frac{4}{3}\right)^2 \approx 1.125 \)
• \( h(1.5) = -(1.5)^3 + 2(1.5)^2 = -3.375 + 4.5 = -1.125 \)

Step 5: Identify the Maximum Value

The highest function value is:

\[ h(-1) = 3 \]

Thus, the maximum value of \( h(x) \) is 3, occurring at \( x = -1 \).