Question

In the interval \([-5,5]\), if \[ f(x)=(x+3)^2(x-2)^3 \] is increasing on \[ S=\{x\mid -5\le x<\alpha \text{ and } \beta<x\le5\}, \] then \(f(\alpha)-f(\beta)=\) ?

• A. \(-108\)
• B. \(108\)
• C. \(72\)
• D. \(-72\)

Hint:

After finding critical points, always make a sign chart for \(f'(x)\) to identify increasing and decreasing intervals.

Answer 1

The Correct Option is B

Concept: A function is increasing where \[ f'(x)>0. \] The critical points are obtained from \[ f'(x)=0. \] These points divide the interval into monotonic regions.

Step 1: Differentiate the function.
Given \[ f(x)=(x+3)^2(x-2)^3. \] Using product rule, \[ f'(x) = 2(x+3)(x-2)^3 + 3(x+3)^2(x-2)^2. \] Taking common factors, \[ f'(x) = (x+3)(x-2)^2 \Big[2(x-2)+3(x+3)\Big]. \] \[ = (x+3)(x-2)^2(5x+5). \] \[ = 5(x+3)(x+1)(x-2)^2. \]

Step 2: Find critical points.
\[ f'(x)=0 \] gives \[ x=-3,\quad x=-1,\quad x=2. \]

Step 3: Determine increasing intervals.
Since \[ (x-2)^2\ge0, \] the sign depends on \[ (x+3)(x+1). \] Positive sign occurs for \[ x<-3 \] and \[ x>-1. \] Hence \[ \alpha=-3, \qquad \beta=-1. \]

Step 4: Evaluate \(f(\alpha)\) and \(f(\beta)\).
\[ f(-3)=0. \] \[ f(-1) = (2)^2(-3)^3 = 4(-27) = -108. \] Therefore \[ f(\alpha)-f(\beta) = 0-(-108) = 108. \] Hence \[ \boxed{108}. \]