Feb/Mar 2021 p12 q7
674
Functions f and g are defined as follows:
\(f : x \mapsto x^2 + 2x + 3\) for \(x \leq -1\),
\(g : x \mapsto 2x + 1\) for \(x \geq -1\).
(a) Express \(f(x)\) in the form \((x+a)^2 + b\) and state the range of \(f\).
(b) Find an expression for \(f^{-1}(x)\).
(c) Solve the equation \(gf(x) = 13\).
Solution
(a) To express \(f(x) = x^2 + 2x + 3\) in the form \((x+a)^2 + b\), complete the square:
\(f(x) = (x+1)^2 + 2\).
The range of \(f\) is \(y \geq 2\) since the vertex of the parabola is at \(y = 2\) and it opens upwards.
(b) To find \(f^{-1}(x)\), start with \(y = (x+1)^2 + 2\).
Solving for \(x\), we have:
\(y - 2 = (x+1)^2\)
\(x+1 = \pm \sqrt{y-2}\)
Since \(x \leq -1\), choose the negative root:
\(x = -\sqrt{y-2} - 1\)
Thus, \(f^{-1}(x) = -\sqrt{x-2} - 1\).
(c) Solve \(gf(x) = 13\):
\(g(f(x)) = 2(x^2 + 2x + 3) + 1 = 13\)
\(2x^2 + 4x + 6 + 1 = 13\)
\(2x^2 + 4x - 6 = 0\)
Divide by 2:
\(x^2 + 2x - 3 = 0\)
Factorize:
\((x-1)(x+3) = 0\)
\(x = 1\) or \(x = -3\)
Since \(x \leq -1\), \(x = -3\) only.
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