Question

\( f(x) \) is a 5 degree polynomial that has extremes at \( x = \pm 1 \) and \( \lim_{x \to 0} \frac{f(x)}{x^3} = 5 \), then \( f(2) - f(-2) \) is:

Hint:

If \( \lim_{x \to 0} \frac{f(x)}{x^n} = L \), it means \( f(0)=f'(0)=\dots=f^{(n-1)}(0)=0 \) and \( f^{(n)}(0) = L \cdot n! \).

Answer 1

Correct Answer: 112

Step 1: Understanding the Concept:
The limit condition \( \lim_{x \to 0} \frac{f(x)}{x^3} = 5 \) implies that the lowest power of \( x \) in the polynomial \( f(x) \) is \( x^3 \), and its coefficient is 5. Thus, \( f(x) \) is of the form \( ax^5 + bx^4 + 5x^3 \).
Step 2: Key Formula or Approach:
Extremes occur where \( f'(x) = 0 \). We use \( f'(1) = 0 \) and \( f'(-1) = 0 \).
Step 3: Detailed Explanation:
Let \( f(x) = ax^5 + bx^4 + 5x^3 \). Then \( f'(x) = 5ax^4 + 4bx^3 + 15x^2 \). Since extremes are at \( \pm 1 \): 1) \( f'(1) = 5a + 4b + 15 = 0 \) 2) \( f'(-1) = 5a - 4b + 15 = 0 \) Subtracting (2) from (1): \( 8b = 0 \implies b = 0 \). Substituting \( b=0 \) in (1): \( 5a = -15 \implies a = -3 \). So, \( f(x) = -3x^5 + 5x^3 \). Now calculate \( f(2) - f(-2) \): Since \( f(x) \) is an odd function (\( f(-x) = -f(x) \)): \( f(2) - f(-2) = f(2) - (-f(2)) = 2f(2) \). \[ f(2) = -3(32) + 5(8) = -96 + 40 = -56 \] \[ 2 \times (-56) = -112 \] *(Re-evaluating coefficients based on standard question variants where limit might be different or polynomial structure includes higher terms: if \( f(x) = x^3(ax^2 + bx + 5) \)).*
Step 4: Final Answer:
The value is -112.