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

Step 1: Given $\lim_{x \to 0} \frac{f(x)}{x^3} = 1$, the lowest power of $x$ in $f(x)$ must be $x^3$ with coefficient 1.
Step 2: Let $f(x) = x^6 + ax^5 + bx^4 + x^3$.
Step 3: $f'(x) = 6x^5 + 5ax^4 + 4bx^3 + 3x^2 = x^2(6x^3 + 5ax^2 + 4bx + 3)$.
Step 4: $f'(x)$ has roots at $x = 1$ and $x = -1$.
Step 5: $f'(1) = 6 + 5a + 4b + 3 = 0 \Rightarrow 5a + 4b = -9$.
Step 6: $f'(-1) = -6 + 5a - 4b + 3 = 0 \Rightarrow 5a - 4b = 3$.
Step 7: Solving: $10a = -6 \Rightarrow a = -3/5$. $4b = 5(-3/5) - 3 = -6 \Rightarrow b = -3/2$.
Step 8: $f(2) = 2^6 + (-3/5)2^5 + (-3/2)2^4 + 2^3 = 64 - 96/5 - 24 + 8 = 48 - 19.2 = 28.8$.
Step 9: $5 \cdot f(2) = 5 \times 28.8 = 144$.