(a) The functions \(f : x \to x^{2} + 1\) and \(g : x \to
(a) The functions \(f : x \to x^{2} + 1\) and \(g : x \to 5 - 3x\) are defined on the set of the real numbers, R.
(i) State the domain of \(f^{-1}\), the inverse of f ; (ii) find \(g^{-1} (2)\).
(b) Evaluate : \(\int \frac{(x + 3)}{x^{2} + 6x + 9} \mathrm {d} x\)
Explanation
(a) \(f : x \to x^{2} + 1\)
\(g : x \to 5 - 3x\)
(i) \(f(x) = x^{2} + 1\)
Let y = f(x).
\(y = x^{2} + 1 \implies x^{2} = y - 1\)
\(x = \sqrt{y - 1}\)
\(\implies f^{-1} (x) = \sqrt{x - 1}\)
\(Dom(f^{-1} (x)) = x \geq 1\)
(ii) \(g(x) = 5 - 3x\)
Let y = g(x)
\(y = 5 - 3x \implies 3x = 5 - y\)
\(x = \frac{5 - y}{3}\)
\(g^{-1} (x) = \frac{5 - x}{3}\)
\(g^{-1} (2) = \frac{5 - 2}{3} = 1\)
(b) \(\int \frac{x + 3}{x^{2} + 6x + 9} \mathrm {d} x\)
\(\frac{x + 3}{x^{2} + 6x + 9} = \frac{x + 3}{(x + 3)^{2}}\)
= \(\frac{1}{x + 3}\)
\(\therefore \int \frac{x + 3}{x^{2} + 6x + 9} \mathrm {d} x = \int \frac{1}{x + 3} \mathrm {d} x\)
= \(\ln |x + 3| + c\)