Key Concepts and Formulas

• Polynomial Limits at Zero: If $\lim_{x \to 0} \frac{P(x)}{x^n} = L \neq 0$, where $P(x)$ is a polynomial, then $P(x)$ has the form $P(x) = Lx^n + a_{n+1}x^{n+1} + ...$, where $L$ is the coefficient of the $x^n$ term. All terms $x^k$ for $k < n$ must have zero coefficients.
• Extrema and Derivatives: If $f(x)$ has a local extremum at $x=c$, and $f(x)$ is differentiable at $x=c$, then $f'(c) = 0$.
• Polynomial Degree and Derivatives: If $f(x)$ is a polynomial of degree $n$, then $f'(x)$ is a polynomial of degree $n-1$.

Step-by-Step Solution

Step 1: Define the general form of the polynomial $f(x)$.

Since $f(x)$ is a polynomial of degree 6 with the coefficient of $x^6$ being 1, we can write it as: $f(x) = x^6 + ax^5 + bx^4 + cx^3 + dx^2 + ex + f$

Step 2: Apply the limit condition $\lim_{x \to 0} \frac{f(x)}{x^3} = 1$.

This condition tells us that the lowest power of $x$ in $f(x)$ with a non-zero coefficient is $x^3$, and its coefficient is 1. Therefore, $f(0) = 0$, $f'(0) = 0$, $f''(0) = 0$, and the coefficient of $x^3$ is 1. This implies $f = 0$, $e = 0$, $d = 0$, and $c = 1$. So, our polynomial becomes: $f(x) = x^6 + ax^5 + bx^4 + x^3$

Step 3: Use the extrema conditions $f'(-1) = 0$ and $f'(1) = 0$.

First, find the derivative of $f(x)$: $f'(x) = 6x^5 + 5ax^4 + 4bx^3 + 3x^2$

Now, apply the conditions $f'(-1) = 0$ and $f'(1) = 0$:

$f'(-1) = 6(-1)^5 + 5a(-1)^4 + 4b(-1)^3 + 3(-1)^2 = -6 + 5a - 4b + 3 = 0 \Rightarrow 5a - 4b = 3$ $f'(1) = 6(1)^5 + 5a(1)^4 + 4b(1)^3 + 3(1)^2 = 6 + 5a + 4b + 3 = 0 \Rightarrow 5a + 4b = -9$

Step 4: Solve the system of equations for $a$ and $b$.

We have the following system of linear equations: $5a - 4b = 3$ $5a + 4b = -9$

Adding the two equations, we get: $10a = -6 \Rightarrow a = -\frac{3}{5}$ Substituting $a = -\frac{3}{5}$ into the second equation: $5\left(-\frac{3}{5}\right) + 4b = -9 \Rightarrow -3 + 4b = -9 \Rightarrow 4b = -6 \Rightarrow b = -\frac{3}{2}$

Step 5: Substitute the values of $a$ and $b$ back into the polynomial $f(x)$.

$f(x) = x^6 - \frac{3}{5}x^5 - \frac{3}{2}x^4 + x^3$

Step 6: Calculate $f(2)$.

$f(2) = (2)^6 - \frac{3}{5}(2)^5 - \frac{3}{2}(2)^4 + (2)^3 = 64 - \frac{3}{5}(32) - \frac{3}{2}(16) + 8 = 72 - \frac{96}{5} - 24 = 48 - \frac{96}{5} = \frac{240 - 96}{5} = \frac{144}{5}$

Step 7: Calculate $5 \cdot f(2)$.

$5 \cdot f(2) = 5 \cdot \frac{144}{5} = 144$

Step 8: Find the value of k such that $144 = 24*k$.

$144 = 24*k$

$k=6$

Common Mistakes & Tips

• Double-check the signs when solving systems of equations. A small error can propagate through the entire solution.
• Remember that extrema imply the derivative is zero at those points.
• When dealing with limits of polynomials, focus on the lowest degree terms.

Summary

We determined the polynomial $f(x)$ by using the given limit condition and the conditions for extrema. The limit condition allowed us to find the coefficients of the lower-degree terms, and the extrema conditions gave us two equations to solve for the remaining coefficients. After finding $f(x)$, we evaluated $f(2)$ and multiplied by 5 to get the final answer.

The final answer is \boxed{6}.