Question

Let \( f(x) \) be a polynomial of degree 6 in \( x \), in which the coefficient of \( x^6 \) is unity and it has extrema at \( x = -1 \) and \( x = 1 \). If \( \displaystyle \lim_{x \to 0} \frac{f(x)}{x^3} = 1 \), then \( 5 \cdot f(2) \) is equal to __________.

Hint:

If $\lim_{x \to 0} \frac{f(x)}{x^n} = k$, then $f(x)$ contains no terms with powers of $x$ smaller than $n$, and the coefficient of $x^n$ is $k$.

Answer 1

Correct Answer: 144

Given the polynomial \( f(x) \) of degree 6, with extrema at \( x = -1 \) and \( x = 1 \), the derivative \( f'(x) \) must have factors \( (x+1) \) and \( (x-1) \). The simplest form for this is \( f'(x) = k(x+1)^m(x-1)^n \). Since \( f(x) \) is degree 6, \( f'(x) \) is degree 5, giving \( m+n=5 \).
To satisfy conditions of extrema, both \( m \) and \( n \) should be odd. A sensible choice is \( m = 3 \) and \( n = 2 \). Thus, \( f'(x) = k(x+1)^3(x-1)^2 \).
Integrating \( f'(x) \) gives:
\( f(x) = \int k(x+1)^3(x-1)^2 \, dx \).
Expanding and integrating, we do:
\( (x+1)^3 = x^3 + 3x^2 + 3x + 1 \) and \( (x-1)^2 = x^2 - 2x + 1 \).
Multiplying these, \( (x+1)^3(x-1)^2 = x^5 - x^3 + x \).
Then \( f(x) = k \int (x^5 - x^3 + x) \, dx = k \left( \frac{x^6}{6} - \frac{x^4}{4} + \frac{x^2}{2} \right) + C \).
The leading coefficient implies \( k(1/6) = 1 \), thus \( k = 6 \). So:
\( f(x) = x^6 - \frac{6}{4}x^4 + 3x^2 + C \).
Limitation \( \lim_{x \to 0} \frac{f(x)}{x^3} = 1 \) gives:
\( \frac{C}{x^3} = 1 \) as \( x \to 0 \), suggesting \( C = 0 \). Thus:
\( f(x) = x^6 - \frac{3}{2}x^4 + 3x^2 \).
Calculating \( 5 \cdot f(2) \):
\( f(2) = 2^6 - \frac{3}{2}(2^4) + 3(2^2) = 64 - 24 + 12 = 52 \).
Thus, \( 5 \cdot f(2) = 5 \cdot 52 = 260 \).