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June 2014 p12 q10
729
Function h is defined by \(h : x \mapsto x^2 + 4x\) for \(x \geq k\), and it is given that h has an inverse.
(v) State the smallest possible value of \(k\).
(vi) Find an expression for \(h^{-1}(x)\).
Solution
To find the smallest possible value of \(k\), we need the function \(h(x) = x^2 + 4x\) to be one-to-one, which occurs when it is either strictly increasing or decreasing. The vertex of the parabola \(x^2 + 4x\) is at \(x = -\frac{b}{2a} = -\frac{4}{2} = -2\). Thus, the smallest possible value of \(k\) is \(-2\) so that the function is increasing for \(x \geq -2\).
To find \(h^{-1}(x)\), start by setting \(y = x^2 + 4x\).
Complete the square: \(y = (x+2)^2 - 4\).
Rearrange to solve for \(x\):
\(y + 4 = (x+2)^2\)
\(\sqrt{y + 4} = x + 2\)
\(x = \sqrt{y + 4} - 2\)
Thus, the inverse function is \(h^{-1}(x) = \sqrt{x+4} - 2\).
