Solving Functional Equation And Limit Problem Find Α + 2β

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This article delves into a fascinating problem involving a functional equation and a limit, ultimately aiming to find the value of α+2β{\alpha + 2\beta}. We are given a function f:R∖{0}→R{f: \mathbb{R} \setminus \{0\} \rightarrow \mathbb{R}} satisfying the equation:

f(x)−6f(1x)=353x−52{ f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2} }

and the limit:

lim⁡x→0(1αx+f(x))=β{ \lim_{x \to 0} \left( \frac{1}{\alpha x} + f(x) \right) = \beta }

where α,β∈R{\alpha, \beta \in \mathbb{R}}. Our goal is to determine the value of α+2β{\alpha + 2\beta}.

H2: Dissecting the Functional Equation

Functional equations can be tricky, but the key often lies in clever substitutions. Our given equation relates f(x){f(x)} and f(1/x){f(1/x)}. To disentangle these, let's substitute x{x} with 1/x{1/x} in the original equation. This yields:

f(1x)−6f(x)=353(1/x)−52=35x3−52{ f\left(\frac{1}{x}\right) - 6f(x) = \frac{35}{3(1/x)} - \frac{5}{2} = \frac{35x}{3} - \frac{5}{2} }

Now we have two equations:

  1. f(x)−6f(1x)=353x−52{ f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2} }
  2. f(1x)−6f(x)=35x3−52{ f\left(\frac{1}{x}\right) - 6f(x) = \frac{35x}{3} - \frac{5}{2} }

This system of equations can be solved for f(x){f(x)}. To eliminate f(1/x){f(1/x)}, we can multiply the second equation by 6 and add it to the first equation. This gives:

f(x)−6f(1x)+6(f(1x)−6f(x))=353x−52+6(35x3−52){ f(x) - 6f\left(\frac{1}{x}\right) + 6\left( f\left(\frac{1}{x}\right) - 6f(x) \right) = \frac{35}{3x} - \frac{5}{2} + 6\left( \frac{35x}{3} - \frac{5}{2} \right) }

Simplifying, we get:

f(x)−36f(x)=353x−52+70x−15{ f(x) - 36f(x) = \frac{35}{3x} - \frac{5}{2} + 70x - 15 }

−35f(x)=353x+70x−352{ -35f(x) = \frac{35}{3x} + 70x - \frac{35}{2} }

Dividing both sides by -35, we obtain an explicit expression for f(x){f(x)}:

f(x)=−13x−2x+12{ f(x) = -\frac{1}{3x} - 2x + \frac{1}{2} }

This is a crucial step. We've now found the explicit form of the function f(x){f(x)}, which will be vital for evaluating the limit.

H2: Tackling the Limit

Now that we have f(x){f(x)}, let's address the limit:

lim⁡x→0(1αx+f(x))=β{ \lim_{x \to 0} \left( \frac{1}{\alpha x} + f(x) \right) = \beta }

Substitute the expression for f(x){f(x)} we found earlier:

lim⁡x→0(1αx−13x−2x+12)=β{ \lim_{x \to 0} \left( \frac{1}{\alpha x} - \frac{1}{3x} - 2x + \frac{1}{2} \right) = \beta }

To evaluate this limit, we need to consider the behavior of each term as x{x} approaches 0. The terms −2x{-2x} and 1/2{1/2} are well-behaved, approaching 0 and 1/2 respectively. However, the terms 1/(αx){1/(\alpha x)} and −1/(3x){-1/(3x)} can cause issues if not handled carefully.

To obtain a finite limit β{\beta}, the terms with 1/x{1/x} must combine to give a finite value. This means the coefficients of 1/x{1/x} must result in a finite value as x{x} approaches 0. Let's combine those terms:

1αx−13x=3−α3αx{ \frac{1}{\alpha x} - \frac{1}{3x} = \frac{3 - \alpha}{3\alpha x} }

For the limit to exist, this term must not diverge as x{x} approaches 0. This happens only if the numerator is zero, implying:

3−α=0{ 3 - \alpha = 0 }

Thus, we find that α=3{\alpha = 3}. This is a critical piece of information.

Now, substitute α=3{\alpha = 3} back into the limit expression:

lim⁡x→0(13x−13x−2x+12)=lim⁡x→0(−2x+12){ \lim_{x \to 0} \left( \frac{1}{3x} - \frac{1}{3x} - 2x + \frac{1}{2} \right) = \lim_{x \to 0} \left( -2x + \frac{1}{2} \right) }

As x{x} approaches 0, the term −2x{-2x} approaches 0, leaving us with:

β=12{ \beta = \frac{1}{2} }

We have successfully determined the values of α{\alpha} and β{\beta}: α=3{\alpha = 3} and β=12{\beta = \frac{1}{2}}.

H2: The Final Calculation: α+2β{\alpha + 2\beta}

Now, we can calculate the value of α+2β{\alpha + 2\beta}:

α+2β=3+2(12)=3+1=4{ \alpha + 2\beta = 3 + 2\left(\frac{1}{2}\right) = 3 + 1 = 4 }

Therefore, the value of α+2β{\alpha + 2\beta} is 4.

H2: Conclusion

We have solved the problem step-by-step, starting with the functional equation, finding an explicit form for f(x){f(x)}, evaluating the limit, and finally calculating α+2β{\alpha + 2\beta}. The key steps involved recognizing the structure of the functional equation, making appropriate substitutions, and carefully analyzing the limit to determine the value of α{\alpha}. The final answer is 4, which corresponds to option (C).

H3: Key Concepts Revisited

  • Functional Equations: These equations define functions implicitly through relationships between function values at different points. Solving them often involves substitutions and algebraic manipulations.
  • Limits: Limits describe the behavior of a function as its input approaches a certain value. Careful consideration of each term's behavior is crucial for evaluating limits, especially when dealing with indeterminate forms.
  • Algebraic Manipulation: Solving the functional equation required careful algebraic steps to isolate f(x){f(x)}. This highlights the importance of a strong foundation in algebra for solving mathematical problems.

This problem demonstrates a powerful combination of techniques from different areas of mathematics. By mastering these concepts, you can tackle a wide range of challenging problems.

H3: Why is this problem important?

This problem is a great example of how different mathematical concepts can be intertwined. It requires a solid understanding of functional equations, limits, and algebraic manipulation. Such problems are often seen in mathematical competitions and serve as excellent exercises for developing problem-solving skills.

Understanding how to solve functional equations is crucial because they appear in various areas of mathematics, including calculus, differential equations, and real analysis. Limits are fundamental to calculus and are used to define continuity, derivatives, and integrals. The algebraic skills honed in solving such problems are universally applicable in mathematics and other scientific disciplines.

Moreover, problems like these encourage logical thinking and a systematic approach to problem-solving, which are valuable skills in any field.

H3: Strategies for Solving Similar Problems

When faced with a problem involving a functional equation and a limit, consider the following strategies:

  1. Functional Equation:

    • Substitution: Look for substitutions that simplify the equation or create a system of equations. In this case, substituting x{x} with 1/x{1/x} was crucial.
    • Look for Patterns: Sometimes, the equation may have a pattern that can be exploited to find a solution.
    • Assume a Form: If possible, assume a specific form for the function (e.g., a polynomial, exponential) and try to determine the coefficients.

  2. Limits:

    • Direct Substitution: First, try direct substitution. If it leads to an indeterminate form (e.g., 0/0, ∞/∞), further analysis is needed.
    • Algebraic Manipulation: Simplify the expression using algebraic techniques like factoring, rationalizing, or combining fractions.
    • L'Hôpital's Rule: If you have an indeterminate form, L'Hôpital's Rule might be applicable (but be cautious and ensure the conditions are met).
    • Consider One-Sided Limits: If the function behaves differently as you approach the limit from the left or right, consider one-sided limits.

  3. Combining Techniques:

    • Solve for the Function: In many cases, finding an explicit form for the function (as we did here) is the key to evaluating the limit.
    • Use the Limit to Constrain the Function: Sometimes, the limit condition can provide information about the function's behavior or parameters.

By applying these strategies, you can enhance your ability to solve complex mathematical problems involving functional equations and limits.