Step-by-step Solution:

Step 1: Understanding the Polynomial Given that \( f(x) \) is a polynomial of degree 4 with extreme values at \( x = 1 \) and \( x = 2 \), we can express \( f(x) \) in the general form: \[ f(x) = ax^4 + bx^3 + cx^2 + dx + e \] --- Step 2: Finding the Derivative To find the extreme values, we need to calculate the derivative \( f'(x) \): \[ f'(x) = 4ax^3 + 3bx^2 + 2cx + d \] Since \( f'(1) = 0 \) and \( f'(2) = 0 \), we substitute these values. --- Step 3: Setting Up the Equations 1. For \( x = 1 \): \[ 4a(1)^3 + 3b(1)^2 + 2c(1) + d = 0 \] \[ 4a + 3b + 2c + d = 0 \quad \text{(Equation 1)} \] 2. For \( x = 2 \): \[ 4a(2)^3 + 3b(2)^2 + 2c(2) + d = 0 \] \[ 32a + 12b + 4c + d = 0 \quad \text{(Equation 2)} \] --- Step 4: Using the Limit Condition We are given: \[ \lim_{x \to 0} \left( 1 + \frac{f(x)}{x^2} \right) = 3 \] This implies: \[ \lim_{x \to 0} \frac{f(x)}{x^2} = 2 \] Thus, we can write: \[ f(x) = 2x^2 + O(x^3) \quad \text{(as \( x \to 0 \))} \] This means that the constant term \( e \) must be zero, and the coefficient of \( x^2 \) is 2. --- Step 5: Updating the Polynomial Now we can update our polynomial: \[ f(x) = ax^4 + bx^3 + 2x^2 + dx \] Since \( d \) must also be zero (to avoid infinite limits), we have: \[ f(x) = ax^4 + bx^3 + 2x^2 \] --- Step 6: Substituting Back into the Derivative Equations Now, substituting \( d = 0 \) and \( e = 0 \) into our equations: 1. From Equation 1: \[ 4a + 3b + 2(2) = 0 \] \[ 4a + 3b + 4 = 0 \quad \text{(Equation 3)} \] 2. From Equation 2: \[ 32a + 12b + 4(2) = 0 \] \[ 32a + 12b + 8 = 0 \quad \text{(Equation 4)} \] --- Step 7: Solving the System of Equations Now we have two equations: 1. \( 4a + 3b + 4 = 0 \) 2. \( 32a + 12b + 8 = 0 \) From Equation 3, we express \( b \): \[ 3b = -4a - 4 \] \[ b = \frac{-4a - 4}{3} \] Substituting this into Equation 4: \[ 32a + 12 \left( \frac{-4a - 4}{3} \right) + 8 = 0 \] Multiplying through by 3 to eliminate the fraction: \[ 96a - 48a - 48 + 24 = 0 \] \[ 48a - 24 = 0 \] \[ a = \frac{1}{2} \] --- Step 8: Finding \( b \) Substituting \( a = \frac{1}{2} \) back into the expression for \( b \): \[ b = \frac{-4\left(\frac{1}{2}\right) - 4}{3} \] \[ b = \frac{-2 - 4}{3} = \frac{-6}{3} = -2 \] --- Step 9: Final Form of the Polynomial Thus, we have: \[ f(x) = \frac{1}{2} x^4 - 2x^3 + 2x^2 \] --- Step 10: Evaluating \( f(2) \) Now we can find \( f(2) \): \[ f(2) = \frac{1}{2} (2^4) - 2(2^3) + 2(2^2) \] \[ = \frac{1}{2} (16) - 2(8) + 2(4) \] \[ = 8 - 16 + 8 \] \[ = 0 \] --- Conclusion Thus, the value of \( f(2) \) is: \[ \boxed{0} \]