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Nov 2012 p12 q2
746
A function \(f\) is such that \(f(x) = \sqrt{\frac{x+3}{2}} + 1\), for \(x \geq -3\). Find
(i) \(f^{-1}(x)\) in the form \(ax^2 + bx + c\), where \(a, b\) and \(c\) are constants,
(ii) the domain of \(f^{-1}\).
Solution
To find the inverse function \(f^{-1}(x)\), we start with the equation:
\(y = \sqrt{\frac{x+3}{2}} + 1\)
Rearrange to make \(x\) the subject:
\(y - 1 = \sqrt{\frac{x+3}{2}}\)
Square both sides:
\((y - 1)^2 = \frac{x+3}{2}\)
Multiply both sides by 2:
\(2(y - 1)^2 = x + 3\)
Expand \((y - 1)^2\):
\(2(y^2 - 2y + 1) = x + 3\)
\(2y^2 - 4y + 2 = x + 3\)
Rearrange to solve for \(x\):
\(x = 2y^2 - 4y + 2 - 3\)
\(x = 2y^2 - 4y - 1\)
Thus, \(f^{-1}(x) = 2x^2 - 4x - 1\).
For the domain of \(f^{-1}\), since \(f(x) = \sqrt{\frac{x+3}{2}} + 1\) is defined for \(x \geq -3\), the range of \(f(x)\) is \(y \geq 1\). Therefore, the domain of \(f^{-1}\) is \(x \geq 1\).
