(a) If (f(x) = frac{x – 3}{2x – 1} , x neq frac{1}{2}) and (g(x) = frac{x – 1}{x + 1}, x neq -1), fing (g circ f).
(b)(i) Sketch the curve (y = 9x – x^{3}) ; (ii) Calculate the total area bounded by the x- axis and the curve (y = 9x – x^{3}).
Explanation
(a) (f(x) = frac{x – 3}{2x – 1}, x neq frac{1}{2})
(g(x) = frac{x – 1}{x + 1}, x neq -1)
(g circ f = g(f(x)) = frac{frac{x – 3}{2x – 1} – 1}{frac{x – 3}{2x – 1} + 1})
= (frac{frac{x – 3 – 2x + 1}{2x – 1}}{frac{x – 3 + 2x – 1}{2x – 1}})
= (frac{x – 3 – 2x + 1}{x – 3 + 2x – 1})
= (frac{- x – 2}{3x – 4}) or (frac{-(x + 2)}{3x – 4}).
(b)(i) (y = 9x – x^{3})
If x = 0, y = 0.
If y = 0, (9x – x^{3} = 0)
(x(9 – x^{2}) = x (3 – x)(3 + x) = 0)
(implies x = 0, -3, 3)
Turning points : (frac{mathrm d y}{mathrm d x} = 9 – 3x^{2})
(9 – 3x^{2} = 0 implies 9 = 3x^{2})
(x^{2} = 3 implies x = pm sqrt{3} = pm 1.73)
When x = 1.73, (y = 9(1.73) – (1.73)^{3} = 15.57 – 5.18 = 10.39)
Point is (1.73, 10.39).
When x = -1.73, (y = 9(-1.73) – (-1.73)^{3} = -15.57 + 5.18 = -10.39)
Point is (-1.73, -10.39).
Maximum or minimum ?
(frac{mathrm d^{2} y}{mathrm d x^{2}} = – 6x^{2})
At x = 1.73, (frac{mathrm d^{2} y}{mathrm d x^{2}} < 0 therefore (1.73, 10.39) text{ is a maximum point})
At x = -1.73, (frac{mathrm d^{2} y}{mathrm d x^{2}} > 0 therefore (-1.73, -10.39) text{ is a maximum point}).
Areas between x = -3 to x = 0 and x = 0 and x = 3 are equal.
(therefore Area = 2 int_{0} ^{3} (9x – x^{3}) mathrm {d} x = 2[frac{9x^{2}}{2} – frac{x^{4}}{4}]_{0} ^{3})
= (81 – 40frac{1}{2} = 40frac{1}{2} sq. units)