f(x) = 2x + 3 1) a) All values of x are the domain Domain = (-∞, ∞) 4 1) b) f(x) = x 2 - 4x - 5 To find the point where the function will be undefined we equate the denominator to zero x 2 - 4x - 5 = 0 b 2 - 4ac 2a -(-4) ± (-4) 2 - 4(1)(-5) x= 2(1) 4 ± 16 + 20 4 ± 36 x= = 2 2 4 ± 6 10 -2 x= = or 2 2 2 x = +5 or - 1 So, the function is undefined at x = +5 and x = -1 x= -b ± All values of x exluding +5 and -1 Domain = (-∞, -1) ∪ (-1, 5) ∪ (5, ∞) 1) c) f(x) = x2 9 - x2 To find the point where the function will be undefined we equate the denominator to zero 9 - x2 = 0 9 - x2 = 02 = 0 x2 = 9 x = ±3 So, the function is undefined at x = +3 and x = -3 All values of x excluding +3 and -3 Domain = (-∞, -3) ∪ (-3, 3) ∪ (3, ∞) 2) a) f(x) = - 4x 2 + 3x + 2 f(x + h) = - 4(x + h) 2 + 3(x + h) + 2 = - 4 x 2 + 2xh + h 2 + 3x + 3h + 2 f(x + h) = - 4x 2 - 8xh - 4h 2 + 3x + 3h + 2 -4x 2 - 8xh - 4h 2 + 3x + 3h + 2 - -4x 2 + 3x + 2 f(x + h) - f(x) = h h f(x + h) - f(x) -4x 2 - 8xh - 4h 2 +3x + 3h+2+4x 2 -3xx-2 = h h f(x + h) - f(x) -8xh - 4h 2 + 3h = h h f(x + h) - f(x) = - 8x - 4h + 3 h (Answer) (Answer) 2) b) f(x) = 1 x2 f(x + h) = 1 (x + h) 2 1 f(x + h) = 2 x + 2xh + h 2 f(x + h) - f(x) = h 1 x 2 +2xh+h 2 - h 1 x2 (Answer) x 2 - x 2 +2xh+h 2 = x 2 -x 2 -2xh-h 2 x 4 +2x 3 h+x 2 h 2 x 2 +2xh+h 2 x 2 h -2xh-h 2 4 x +2x 3 h+x 2 h 2 f(x + h) - f(x) = = h h h 2 f(x + h) - f(x) -2xh - h h(-2x - h) 1 = 4 ÷ h = × h x + 2x 3 h + x 2 h 2 x 4 + ...
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