- Identify the excluded values for the domains: \(x
eq -9\) for \(f(x)\) and \(x
eq 7\) for \(g(x)\).
- Show the composition \(f(g(x)) = \frac{7(\frac{9x-1}{7-x}) + 1}{(\frac{9x-1}{7-x}) + 9} = \frac{62x}{62} = x\).
- Show the composition \(g(f(x)) = \frac{9(\frac{7x+1}{x+9}) - 1}{7 - (\frac{7x+1}{x+9})} = \frac{62x}{62} = x\).
- Conclude that \(f\) and \(g\) are inverses since both compositions return the identity value \(x\).
Explanation
- Determining the Domains of the Functions
To verify that \(f(x) = \frac{7x+1}{x+9}\) and \(g(x) = \frac{9x-1}{7-x}\) are inverses, we must first define their domains by identifying any values that make their respective denominators equal to zero.
For the function \(f(x) = \frac{7x+1}{x+9}\), the denominator is \(x+9\). Setting this equal to zero, we find \(x + 9 = 0\), which implies \(x = -9\). Therefore, the domain of \(f\) is all real numbers except \(x = -9\).
For the function \(g(x) = \frac{9x-1}{7-x}\), the denominator is \(7-x\). Setting this equal to zero, we find \(7 - x = 0\), which implies \(x = 7\). Therefore, the domain of \(g\) is all real numbers except \(x = 7\).
- Verifying f(g(x)) = x
To show that \(f\) and \(g\) are inverses, we must demonstrate that \(f(g(x)) = x\) and \(g(f(x)) = x\).
First, let us calculate the composition \(f(g(x))\):
\[f(g(x)) = \frac{7\left(\frac{9x-1}{7-x}\right) + 1}{\left(\frac{9x-1}{7-x}\right) + 9}\]
To simplify the numerator:
\[7\left(\frac{9x-1}{7-x}\right) + 1 = \frac{63x - 7}{7-x} + \frac{7-x}{7-x} = \frac{62x}{7-x}\]
To simplify the denominator:
\[\frac{9x-1}{7-x} + 9 = \frac{9x-1}{7-x} + \frac{9(7-x)}{7-x} = \frac{9x-1+63-9x}{7-x} = \frac{62}{7-x}\]
Dividing the numerator by the denominator:
\[f(g(x)) = \frac{62x}{7-x} \div \frac{62}{7-x} = \frac{62x}{7-x} \cdot \frac{7-x}{62} = x\]
This holds for all \(x\) in the domain of \(g\), i.e., \(x
eq 7\).
- Verifying g(f(x)) = x and Concluding the Proof
Next, let us calculate the composition \(g(f(x))\):
\[g(f(x)) = \frac{9\left(\frac{7x+1}{x+9}\right) - 1}{7 - \left(\frac{7x+1}{x+9}\right)}\]
To simplify the numerator:
\[9\left(\frac{7x+1}{x+9}\right) - 1 = \frac{63x+9}{x+9} - \frac{x+9}{x+9} = \frac{62x}{x+9}\]
To simplify the denominator:
\[7 - \left(\frac{7x+1}{x+9}\right) = \frac{7(x+9)}{x+9} - \frac{7x+1}{x+9} = \frac{7x+63-7x-1}{x+9} = \frac{62}{x+9}\]
Dividing the numerator by the denominator:
\[g(f(x)) = \frac{62x}{x+9} \div \frac{62}{x+9} = \frac{62x}{x+9} \cdot \frac{x+9}{62} = x\]
This holds for all \(x\) in the domain of \(f\), i.e., \(x
eq -9\). Since \(f(g(x)) = x\) and \(g(f(x)) = x\), we conclude that the functions are inverses of each other.
Examples
Inverse functions are fundamental in fields like cryptography and data processing. For instance, if a specific function \(f(x)\) is used to encrypt data (transforming a plain-text message \(x\) into cipher-text \(y\)), the inverse function \(g(y)\) is essential to decrypt the cipher-text and recover the original message. Verifying that two functions are inverses ensures that the decryption process perfectly reverses the encryption process, maintaining data integrity and security.