(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)