Chapter 12: Problem 51
Let \(\mathrm{f}(\mathrm{x})\) be a polynomial of degree four having extreme values at \(\mathrm{x}=1\) and \(\mathrm{x}=\) 2. If \(\lim _{\mathrm{x} \rightarrow 0}\left[1+\frac{\mathrm{f}(\mathrm{x})}{\mathrm{x}^{2}}\right]=3\), then \(\mathrm{f}(2)\) is equal to : (a) 0 (b) 4 (c) \(-8\) (d) \(-4\)
Short Answer
Step by step solution
Understand the Nature of Derivative
Consider the Limit Condition
Algebraic Formulation
Solve for Specific \( f(x) \)
Solve Result from Steps into Point \( x=2\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Extreme Values
For a polynomial function like \( f(x) \) of degree 4, we know that it can have up to three extreme points. These occur where \( f'(x) = 0 \). This means if a polynomial \( f(x) \) has extreme values at \( x = 1 \) and \( x = 2 \), then the derivative \( f'(x) \) must have factors \((x - 1) \) and \((x - 2) \).
- The polynomial is at its maximum or minimum at these points.
- For a degree 4 polynomial, the derivative \( f'(x) \) has to be a degree 3 polynomial.
- It takes the form \( f'(x) = (x-1)(x-2)g(x) \), where \( g(x) \) is a second degree polynomial.
Limit of a Function
In this polynomial exercise, we're tasked with solving \( \lim_{x \to 0} \left[1 + \frac{f(x)}{x^2}\right] = 3 \). This tells us how \( f(x) \) behaves as \( x \) gets very small. To satisfy this limit condition, we express \( f(x) \, \) such that it includes \( x^2(h(x) + 2) \), where \( h(x) \) is another polynomial.
- The structure \( x^2(h(x) + 2) \) aligns with the limit condition turning into a point of convergence at 3.
- This condition helps in constructing \( f(x) \) in a form that balances both derivative properties and limit approaches.
- It assists in pinpointing unforeseen values of the polynomial when traditional formulas and direct calculations fall short.
Polynomial Derivatives
For a fourth degree polynomial like \( f(x) \), the derivative \( f'(x) \) would be a third degree polynomial. Since we know the extreme points \( x = 1 \) and \( x = 2 \) make \( f'(x) = 0 \) at those points, we can represent the derivative as:
- \( f'(x) = (x-1)(x-2)g(x) \) where \( g(x) \) is a degree 2 polynomial.
- This product equals zero at \( x = 1 \) and \( x = 2 \), as required for the extreme values.
By solving through these lens, one can determine values such as \( f(2) = -8 \) confidently once all polynomial and derivative conditions are aptly considered.