Question

Let \( f(x) \) be a polynomial of degree 5, and have extrema at \( x = 1 \) and \( x = -1 \). If \( \lim_{x \to 0} \frac{f(x)}{x^3} = -5 \), then \( f(2) - f(-2) \) is equal to:

• A. 0
• B. 50
• C. 92
• D. 112

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The limit condition tells us about the lowest degree terms of the polynomial.
For the limit \(\lim_{x \to 0} \frac{f(x)}{x^3}\) to exist and equal a non-zero finite value, the polynomial \(f(x)\) must not have any \(x^0\), \(x^1\), or \(x^2\) terms.
The extrema conditions tell us that the derivative \(f'(x)\) evaluates to zero at specific points.
Step 2: Key Formula or Approach:
Let the 5th degree polynomial be \(f(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + f\).
Due to the limit condition \(\lim_{x \to 0} \frac{f(x)}{x^3} = -5\), we must have \(f = 0\), \(e = 0\), \(d = 0\), and \(c = -5\).
So, \(f(x) = ax^5 + bx^4 - 5x^3\).
For extrema at \(x = \pm 1\), we use \(f'(1) = 0\) and \(f'(-1) = 0\).
Step 3: Detailed Explanation:
Find the first derivative of the simplified polynomial.
\[ f'(x) = 5ax^4 + 4bx^3 - 15x^2 \] Apply the condition \(f'(1) = 0\).
\[ 5a(1)^4 + 4b(1)^3 - 15(1)^2 = 0 \implies 5a + 4b - 15 = 0 \quad \text{--- (Eq 1)} \] Apply the condition \(f'(-1) = 0\).
\[ 5a(-1)^4 + 4b(-1)^3 - 15(-1)^2 = 0 \implies 5a - 4b - 15 = 0 \quad \text{--- (Eq 2)} \] Add (Eq 1) and (Eq 2) to eliminate \(b\).
\[ 10a - 30 = 0 \implies a = 3 \] Subtract (Eq 2) from (Eq 1) to find \(b\).
\[ 8b = 0 \implies b = 0 \] Now we have the full equation for the polynomial.
\[ f(x) = 3x^5 - 5x^3 \] We need to calculate \(f(2) - f(-2)\).
First, calculate \(f(2)\).
\[ f(2) = 3(2)^5 - 5(2)^3 = 3(32) - 5(8) = 96 - 40 = 56 \] Since \(f(x)\) only has odd powers, it is an odd function, meaning \(f(-x) = -f(x)\).
So, \(f(-2) = -f(2) = -56\).
Finally, compute the difference.
\[ f(2) - f(-2) = 56 - (-56) = 112 \] Step 4: Final Answer:
The value of \(f(2) - f(-2)\) is \(112\).