Functions f and g are defined by
\(f : x \mapsto 2x - 5, \; x \in \mathbb{R},\)
\(g : x \mapsto \frac{4}{2-x}, \; x \in \mathbb{R}, \; x \neq 2.\)
(ii) Express each of \(f^{-1}(x)\) and \(g^{-1}(x)\) in terms of \(x\).
(iii) Show that the equation \(f^{-1}(x) = g^{-1}(x)\) has no real roots.
Solution
(ii) To find \(f^{-1}(x)\), start with \(y = 2x - 5\). Solve for \(x\):
\(y + 5 = 2x\)
\(x = \frac{1}{2}(y + 5)\)
Thus, \(f^{-1}(x) = \frac{1}{2}(x + 5)\).
To find \(g^{-1}(x)\), start with \(y = \frac{4}{2-x}\). Solve for \(x\):
\(y(2-x) = 4\)
\(2y - xy = 4\)
\(xy = 2y - 4\)
\(x = \frac{2y - 4}{y}\)
\(x = 2 - \frac{4}{y}\)
Thus, \(g^{-1}(x) = 2 - \frac{4}{x}\).
(iii) Set \(f^{-1}(x) = g^{-1}(x)\):
\(\frac{1}{2}(x + 5) = 2 - \frac{4}{x}\)
Multiply through by 2x to clear fractions:
\(x(x + 5) = 4x - 8\)
\(x^2 + 5x = 4x - 8\)
\(x^2 + x + 8 = 0\)
Use the discriminant \(b^2 - 4ac\):
\(1^2 - 4 \times 1 \times 8 = 1 - 32 = -31\)
Since the discriminant is negative, there are no real roots.
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