Question

6. Which of the following function has an inverse function (a) \( f(x)=\frac{1}{x-1} \) (b) \( f(x)=x^{2} \) for all \( x \) (c) \( f(x)=x^{2}, x \geq 0 \) (d) \( f(x)=x^{2}, x \leq 0 \)

Answer 1

\(f(x) = \frac{1}{x-1}\)

\(y = \frac{1}{x-1}\)

yx - y = 1

yx = 1 + y

\(x = \frac{1+y}{y}\)  Replace \(y \leftrightarrow x\)   

\(\frac{y=\frac{1+x}{x}}{f(x) = \frac{1+x}{x}}\)  

\(y = x^2\)  for all x

\(x = \sqrt y\)

Replace \(y \leftrightarrow x\)

\(y = \sqrt x\)   \(x \geq 0\)  so not true for all x

\(y = \sqrt x\) 

Answer 2

(a) \(f(x)= \frac{1}{x - 1}\):
This is one-to-one because for every input \(x\), there is a unique output. It passes the horizontal line test.
Has an inverse.

(b) \(f(x)= x^2\), for all \(x\):
This is not one-to-one because \(x^2=(-x)^2\) (e.g., \(f(2)=f(-2)=4\)).
Does not have an inverse.

(c) \(f(x)= x^2\), \(x \geq 0\):
This restriction makes the function one-to-one because negative values are excluded. For \(x \geq 0, x^2\) is unique for each \(x\).
Has an inverse.

(d) \(f(x)= x^2\), \(x \leq 0\):
Similarly, restricting \(x \leq 0\) makes the function one-to-one because positive values are excluded. For \(x \leq 0, x^2\) is unique for each \(x\).
Has an inverse.

Correct Options:

• (a), (c), (d)