Question

The function \( f(x) = (x - 4)^2 (1 + x)^3 \) attains a local extremum at the point:

• A. \( x = 2 \)
• B. \( x = -1 \)
• C. \( x = 0 \)
• D. \( x = 1 \)
• E. \( x = -2 \)

Hint:

When finding critical points of a function that is a product of two terms, use the product rule to differentiate and factorize the expression to solve for the critical points.

Answer 1

The Correct Option is A

Step 1: To find the critical points of the function, we first calculate its derivative. Start by applying the product rule: \[ f'(x) = \frac{d}{dx} \left[ (x - 4)^2 (1 + x)^3 \right]. \] By the product rule: \[ f'(x) = 2(x - 4)(1 + x)^3 + (x - 4)^2 \cdot 3(1 + x)^2. \] Step 2: Factor out the common terms from the derivative expression: \[ f'(x) = (x - 4)(1 + x)^2 \left[ 2(1 + x) + 3(x - 4) \right]. \] Step 3: Simplify the expression inside the brackets: \[ 2(1 + x) + 3(x - 4) = 2 + 2x + 3x - 12 = 5x - 10. \] Thus, the derivative becomes: \[ f'(x) = (x - 4)(1 + x)^2 (5x - 10). \] Step 4: To find the critical points, set the derivative equal to zero: \[ (x - 4)(1 + x)^2 (5x - 10) = 0. \] Step 5: Solve the equation: - \( x - 4 = 0 \) gives \( x = 4 \), - \( (1 + x)^2 = 0 \) gives \( x = -1 \), - \( 5x - 10 = 0 \) gives \( x = 2 \). 
Thus, the critical points are \( x = 4 \), \( x = -1 \), and \( x = 2 \). 
Step 6: To determine whether these points are local extrema, check the second derivative or use the first derivative test. 
By evaluating the function behavior or using the second derivative test, we find that \( x = 2 \) corresponds to a local extremum.