Question

Find a formula for the inverse of the function.

24. \(g\left( x \right) = {x^2} - 2x,x \ge 1\)

Short answer

The inverse of the function is \({g^{ - 1}}\left( x \right) = 1 + \sqrt {x + 1} \).

Step-by-step solution

Condition to find the inverse function of a one-to-one function

Step 1: Write the equation as \(y = f\left( x \right)\).

Step 2: If possible solve the equation in terms of \(y\).

Step 3: Express the inverse \({f^{ - 1}}\) as a function of \(x\), for that interchange\(x\) and \(y\). The equation becomes \(y = {f^{ - 1}}\left( x \right)\).

Determine the formula for the inverse of the function 

Complete the square of the equation as shown below:

\(\begin{aligned}{x^2} - 2x &= \left( {{x^2} - 2x + 1} \right) - 1\\ &= {\left( {x - 1} \right)^2} - 1\end{aligned}\)

The restriction is \(x \ge 1\).

It is observed that \(g\) is a one-to-one function.

Write the obtained equation as \(y = {\left( {x - 1} \right)^2} - 1,x \ge 1\) and solve the equation for \(x\) as shown below:

\(\begin{aligned}y &= {\left( {x - 1} \right)^2} - 1\\y + 1 &= {\left( {x - 1} \right)^2}\\x - 1 &= \sqrt {y + 1} \left( {{\mathop{\rm since}\nolimits} \,\,x \ge 1 \Leftrightarrow x - 1 \ge 0} \right)\\x &= 1 + \sqrt {y + 1} \end{aligned}\)

Interchange \(x\) and \(y\) in the above equation as shown below:

\(y = 1 + \sqrt {x + 1} \)

Thus, the inverse of the function is \({g^{ - 1}}\left( x \right) = 1 + \sqrt {x + 1} \).