Question

If \(f(x) = \frac{2x-1}{x+5}\), \(x\neq -5\) then \(f^{-1}(x)\) is equal to

• A. \(\frac{x+5}{2x-1}\), \(x\neq \frac{1}{2}\)
• B. \(\frac{5x+1}{2-x}\), \(x\neq 2\)
• C. \(\frac{5x-1}{2-x}\), \(x\neq 2\)
• D. \(\frac{x-5}{2x+1}\), \(x\neq -\frac{1}{2}\)

Hint:

Put y = f(x), solve for x, then swap the letters.

Answer 1

The Correct Option is B

Step 1: Approach
Verify by composing $f$ with each candidate inverse and looking for $x$.

Step 2: Test option (B)
Let $g(x)=\dfrac{5x+1}{2-x}$. Then
\[ f(g(x))=\frac{2g-1}{g+5}=\frac{\frac{10x+2-2+x}{2-x}}{\frac{5x+1+10-5x}{2-x}}=\frac{11x}{11}=x \]

Step 3: Conclusion
$f(g(x))=x$, so $g=f^{-1}$, and the restriction $x\ne2$ comes from the zero of the denominator. Option (B) is correct.

Final Answer:
The inverse function is $\dfrac{5x+1}{2-x}$ with $x\ne2$, option (B). \[ \boxed{\frac{5x+1}{2-x},\ x\ne2} \]