Key Concepts and Formulas

• Extrema of a Polynomial: A polynomial $f(x)$ has an extremum (local maximum or minimum) at a point $x=c$ if $f'(c) = 0$, provided $f'(x)$ exists at $x=c$.
• Limits of Polynomials: For a polynomial $P(x)$, $\lim_{x \to a} P(x) = P(a)$. When dealing with limits involving fractions of polynomials where the denominator approaches zero, the numerator must also approach zero for the limit to be finite. This often implies the presence of common factors.
• Polynomial Structure: A polynomial of degree $n$ can be represented as $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, where $a_i$ are coefficients and $a_n \neq 0$.

Step-by-Step Solution

Step 1: Understand the conditions given in the problem. We are given that $f(x)$ is a polynomial of degree 4. It has extreme values at $x=1$ and $x=2$. This implies that the derivative of $f(x)$, denoted by $f'(x)$, must be zero at these points.

• $f'(1) = 0$
• $f'(2) = 0$

We are also given a limit condition: $\mathop {lim}\limits_{x \to 0} \left( {{{f\left( x \right)} \over {{x^2}}} + 1} \right) = 3$.

Step 2: Represent the polynomial $f(x)$ and its derivative $f'(x)$. Since $f(x)$ is a polynomial of degree 4, we can write it in the general form: $f(x) = Ax^4 + Bx^3 + Cx^2 + Dx + E$ where $A, B, C, D, E$ are constants and $A \neq 0$.

The derivative of $f(x)$ is: $f'(x) = 4Ax^3 + 3Bx^2 + 2Cx + D$

Step 3: Use the limit condition to find information about the coefficients. The given limit is: $\mathop {lim}\limits_{x \to 0} \left( {{{f\left( x \right)} \over {{x^2}}} + 1} \right) = 3$ Substitute the general form of $f(x)$: $\mathop {lim}\limits_{x \to 0} \left( {{{Ax^4 + Bx^3 + Cx^2 + Dx + E} \over {{x^2}}} + 1} \right) = 3$ Separate the terms in the fraction: $\mathop {lim}\limits_{x \to 0} \left( {Ax^2 + Bx + C + \frac{D}{x} + \frac{E}{x^2} + 1} \right) = 3$ For this limit to exist and be finite, the terms with $x$ in the denominator must vanish. This means the coefficients of these terms must be zero. Therefore, $D = 0$ and $E = 0$.

Now, the limit becomes: $\mathop {lim}\limits_{x \to 0} \left( {Ax^2 + Bx + C + 1} \right) = 3$ Since $Ax^2 + Bx + C + 1$ is a polynomial in $x$, we can evaluate the limit by direct substitution: $A(0)^2 + B(0) + C + 1 = 3$ $C + 1 = 3$ $C = 2$

So far, we have $f(x) = Ax^4 + Bx^3 + 2x^2$, and $f'(x) = 4Ax^3 + 3Bx^2 + 4x$.

Step 4: Use the conditions on the derivative to form equations. We know that $f'(1) = 0$ and $f'(2) = 0$. Using $f'(x) = 4Ax^3 + 3Bx^2 + 4x$:

For $f'(1) = 0$: $4A(1)^3 + 3B(1)^2 + 4(1) = 0$ $4A + 3B + 4 = 0$ $4A + 3B = -4 \quad (*)$

For $f'(2) = 0$: $4A(2)^3 + 3B(2)^2 + 4(2) = 0$ $4A(8) + 3B(4) + 8 = 0$ $32A + 12B + 8 = 0$ Divide by 4: $8A + 3B + 2 = 0$ $8A + 3B = -2 \quad (**)$

Step 5: Solve the system of linear equations for $A$ and $B$. We have two equations with two unknowns:

1. $4A + 3B = -4$
2. $8A + 3B = -2$

Subtract equation (1) from equation (2): $(8A + 3B) - (4A + 3B) = -2 - (-4)$ $4A = 2$ $A = \frac{2}{4} = \frac{1}{2}$

Substitute the value of $A$ into equation (1): $4\left(\frac{1}{2}\right) + 3B = -4$ $2 + 3B = -4$ $3B = -6$ $B = -2$

Step 6: Write the complete expression for $f(x)$ and calculate $f(-1)$. Now we have all the coefficients: $A = \frac{1}{2}$, $B = -2$, $C = 2$, $D = 0$, $E = 0$. So, the polynomial $f(x)$ is: $f(x) = \frac{1}{2}x^4 - 2x^3 + 2x^2$

We need to find $f(-1)$: $f(-1) = \frac{1}{2}(-1)^4 - 2(-1)^3 + 2(-1)^2$ $f(-1) = \frac{1}{2}(1) - 2(-1) + 2(1)$ $f(-1) = \frac{1}{2} + 2 + 2$ $f(-1) = \frac{1}{2} + 4$ $f(-1) = \frac{1}{2} + \frac{8}{2}$ $f(-1) = \frac{9}{2}$

Common Mistakes & Tips

• Forgetting the condition for finite limits: When a limit of a rational function results in an indeterminate form like $\frac{0}{0}$ or $\frac{k}{0}$, if the limit is finite, the numerator must have factors that cancel out the terms causing the denominator to be zero. In this case, $\frac{f(x)}{x^2}$ as $x \to 0$ requires $f(0)=0$ and $f'(0)=0$ for the limit to be finite. This translates to $E=0$ and $D=0$.
• Algebraic errors in solving simultaneous equations: Double-check the calculations when solving for the coefficients $A$ and $B$. A small error here will propagate to the final answer.
• Incorrectly differentiating the polynomial: Ensure the power rule for differentiation is applied correctly. The derivative of $Ax^n$ is $nAx^{n-1}$.

Summary

The problem involves finding a degree 4 polynomial $f(x)$ given conditions about its extrema and a limit. The conditions for extrema at $x=1$ and $x=2$ translate to $f'(1)=0$ and $f'(2)=0$. The limit condition as $x \to 0$ for $f(x)/x^2$ being finite implies that the constant term and the coefficient of the $x$ term in $f(x)$ must be zero. By setting up and solving a system of linear equations for the coefficients of $f(x)$, we determine the polynomial and can then evaluate $f(-1)$.

The final answer is $\boxed{{9 \over 2}}$.