Question

11 mark for review let ( f ) be a function with first derivative given by ( f(x) = (x + 1)(x - 2)(x - 3) ). at what values of ( x ) does ( f ) have a relative maximum? a 2 only b ( -1 ) and 3 only c 3 only d ( -1, 2, ) and 3

Explanation:

Step1: Find critical points

To find critical points, set \( f'(x) = 0 \). Given \( f'(x)=(x + 1)(x - 2)(x - 3) \), we solve \((x + 1)(x - 2)(x - 3)=0\). The solutions are \( x=-1 \), \( x = 2 \), and \( x=3 \). These are the critical points.

Step2: Analyze sign changes of \( f'(x) \)

We use test intervals around the critical points: \( (-\infty, -1) \), \( (-1, 2) \), \( (2, 3) \), and \( (3, \infty) \).

• For \( x < -1 \) (e.g., \( x=-2 \)): \( (-2 + 1)(-2 - 2)(-2 - 3)=(-1)(-4)(-5)=-20<0 \), so \( f'(x)<0 \).
• For \( -1 < x < 2 \) (e.g., \( x = 0 \)): \( (0 + 1)(0 - 2)(0 - 3)=(1)(-2)(-3)=6>0 \), so \( f'(x)>0 \).
• For \( 2 < x < 3 \) (e.g., \( x = 2.5 \)): \( (2.5 + 1)(2.5 - 2)(2.5 - 3)=(3.5)(0.5)(-0.5)=-0.875<0 \), so \( f'(x)<0 \).
• For \( x > 3 \) (e.g., \( x = 4 \)): \( (4 + 1)(4 - 2)(4 - 3)=(5)(2)(1)=10>0 \), so \( f'(x)>0 \).

A relative maximum occurs where \( f'(x) \) changes from positive to negative. At \( x = 2 \), \( f'(x) \) changes from positive (in \( (-1, 2) \)) to negative (in \( (2, 3) \)). At \( x=-1 \), \( f'(x) \) changes from negative to positive (relative minimum). At \( x = 3 \), \( f'(x) \) changes from negative to positive (relative minimum).

Answer:

A. 2 only