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Chapter 1: Problem 58

Find the value of a so that \(f(x)=\frac{a x+1}{x+3}\) is identical with \(f^{-1}(x)\)

Short Answer

Expert verified
After solving the equation for 'a', we find that the variable 'a' can take any real number value except -3. Therefore, the function \(f(x)=\frac{ax+1}{x+3}\) is identical with its inverse \(f^{-1}(x)\) for all 'a' belonging to Real numbers except -3.

Step by step solution

01

Writing the functions

To start, we write down the original function \(f(x) = \frac{ax+1}{x+3}\) and rename \(f^{-1}(x)\) as \(y\), which gives \(y = \frac{ax+1}{x+3}\).
02

Finding the inverse function

To find the inverse function, we switch the variables of our original function. This gives us \(x = \frac{ay+1}{y+3}\). Now we solve this for 'y' to obtain the inverse function \(f^{-1}(x)\). This is done by cross multiplying to remove the fraction, rearranging terms, and then isolating 'y'.
03

Equating the original and inverse functions

Now, since we are told that the original function is identical with its inverse, we equate \(f(x)\) with \(f^{-1}(x)\), substitute the expressions obtained in step 1 and 2 into this equality, and solve for 'a'.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Identical Functions
Identical functions are an interesting concept in mathematics. They refer to two functions being precisely the same or identical across their domains.
In the context of the problem, we were given two functions: the original function \( f(x) = \frac{ax + 1}{x + 3} \) and its inverse \( f^{-1}(x) \). The challenge was to find the value of 'a' which makes \( f(x) = f^{-1}(x) \) for all values of \( x \) in the domain where the functions are defined.
  • When functions are identical, they would yield the same output for any given input value.
  • This means they intersect and overlap perfectly on a graph throughout their range and domain.
  • Identical functions share every characteristic and point of the other.
Achieving this requires solving for 'a' under the condition that the function and its inverse are equal.
Cross Multiplication
Cross multiplication is a method used to solve equations that involve fractions by eliminating the denominator.
It's a powerful tool for simplifying proportions or fractional equations to facilitate further solving. In this exercise:
  • The function \( y = \frac{ax + 1}{x + 3} \) was rewritten by switching 'x' and 'y'. This resulted in the equation \( x = \frac{ay + 1}{y + 3} \).
  • To express this without fractions, you use cross multiplication. This means multiplying diagonally across the equation, which transforms the expression into \( x(y + 3) = ay + 1 \).
  • This step allowed us to remove the fraction and work towards isolating 'y', which is crucial for finding the inverse function.
Cross multiplication is essential in managing complex equations involving fractions and simplifying into solvable chunks.
Solving Equations
Solving equations is a fundamental skill that allows us to find unknown variables that satisfy a mathematical statement.
In this problem, solving equations involved several steps:
  • The equation derived from cross multiplication: \( x(y + 3) = ay + 1 \) was manipulated to isolate 'y'.
  • First, expand \( x(y + 3) \) to \( xy + 3x \), giving \( xy + 3x = ay + 1 \).
  • Bring like terms together, yielding \( xy - ay = 1 - 3x \).
  • Factor out 'y' to isolate it: \( y(x - a) = 1 - 3x \).
  • Solve for 'y': \( y = \frac{1 - 3x}{x - a} \).
This equation provides the inverse function \( f^{-1}(x) \). Solving equationally is strategic as it breaks down problems and leads to clear solutions.
Function Composition
Function composition involves combining two functions to form a new function.
It allows us to explore relationships between functions, especially useful in verifying our work. For identical functions, composing a function with its inverse should return the input value:
  • If \( f \) and \( f^{-1} \) are composed, the result \( f(f^{-1}(x)) \) should equal \( x \).
  • This is known as the identity function for that particular composition: \( f(f^{-1}(x)) = x \).
  • Likewise, \( f^{-1}(f(x)) = x \).
In this problem, both function and its inverse required to generate \( x \), which validated finding the correct 'a.'
Understanding composition is essential for confirming solutions and checking the correctness of inverse calculations.

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Most popular questions from this chapter

If \(f(x)\) is a polynomial of degree 2 such that \(f(0)=1\) and \(f(x+2)=f(x)+4 x+2\), find the polynomial \(f(x)\).

Find the domains of \(f(x)=\sin ^{-1}\left(\log _{2} x\right)\).

Find the domain of \(f(x)=\cos ^{-1}\left(\frac{3}{4+2 \sin x}\right)\)

Find \(f_{o} g\) and \(g_{o} f\) for the functions \(f(x)=\sin x\) and \(g(x)=\sqrt{x-2}\)

The domain of the function \(f(x)=\operatorname{logs}_{(4-x)}(x-1)-\sin ^{-1}[2 x-3]\) is (a) \((1,2)\) (b) \((1,2.5)\) (c) \((1,1.5)\) (d) \((3,4)\).
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