Question

Local maximum and local minimum values of \( f(x) = (x-1)^2(x+2) \) are attained respectively at

• A. \( x = 1 \) and \( x = -1 \)
• B. \( x = -1 \) and \( x = 1 \)
• C. \( x = 1 \) and \( x = 2 \)
• D. \( x = 2 \) and \( x = 1 \)

Hint:

Use the first and second derivative tests to find critical points and determine if they correspond to local maxima or minima.

Answer 1

The Correct Option is A

Step 1: Differentiating the function.
The function is \( f(x) = (x-1)^2(x+2) \). To find the critical points, we differentiate \( f(x) \): \[ f'(x) = 2(x-1)(x+2) + (x-1)^2 \] Simplifying: \[ f'(x) = (x-1)[2(x+2) + (x-1)] = (x-1)(3x+3) \] Thus, \( f'(x) = (x-1)(3x+3) \).
Step 2: Solving for critical points.
Set \( f'(x) = 0 \): \[ (x-1)(3x+3) = 0 \] This gives two solutions: \[ x-1 = 0 \quad \Rightarrow \quad x = 1 \] \[ 3x+3 = 0 \quad \Rightarrow \quad x = -1 \]
Step 3: Second derivative test.
To determine whether these points are maxima or minima, we compute the second derivative: \[ f''(x) = 6(x+2) + 3(x-1) = 9x + 15 \] Evaluating at \( x = 1 \): \[ f''(1) = 9(1) + 15 = 24 \quad (\text{positive, so a local minimum}) \] Evaluating at \( x = -1 \): \[ f''(-1) = 9(-1) + 15 = 6 \quad (\text{positive, so a local maximum}) \]
Step 4: Conclusion.
Therefore, the local maximum occurs at \( x = -1 \) and the local minimum occurs at \( x = 1 \). The correct answer is (1) \( x = 1 \) and \( x = -1 \).