G(x) = X/(x+3) + 3(2x+1)/(x^2 +x - 6) A)Show That G(x) =(x+1 ... - MyTutor

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Log inSign upAnswers>Maths>A Level>Articleg(x) = x/(x+3) + 3(2x+1)/(x^2 +x - 6) a)Show that g(x) =(x+1)/(x-2), x>3 b)Find the range of g c)Find the exact value of a for which g(a)=g^(-1)(a).

a) First, we need to factorize (x2+x-6) = (x-2)(x+3). Therefore, g(x) = x/(x+3) + 3(2x+1)/(x-2)(x+3) = x(x-2)/(x-2)(x+3) + 3(2x+1)/(x-2)(x+3) = (x2-2x + 6x +3)/(x-2)(x+3) = (x2+4x+3)/(x-2)(x+3).Now we need to factorize x2+4x+3 = (x+1)(x+3) => g(x) = (x+1)(x+3)/(x-2)(x+3) => g(x) = (x+1) /(x-2).b) g is a decreasing function, therefore for any x> 3 , g(x)< g(3) = 4/1 = 4. Also, g(x)< lim (g(x)) when x goes to infinity => g(x)< 1Therefore the range of g is the interval (1,4).c) Firstly, we need to compute g^(-1)(x). Let y = (x+1)/(x-2) => y(x-2) = x+1 => yx-x = 2y + 1 => x(y-1)= 2y+1 => x = (2y+1)/(y-1).therefore, g^(-1)(y) = (2y+1)/(y-1). We need to find a such that (a+1)/(a-2) = (2a+1)/(a-1) => a2-1 = 2a2-3a -2 => a2-3a-1= 0=> a= (3 +- sqrt(9+4))/2 = (3 +- sqrt(13))/2 However, a needs to be in the range of g, so the inverse can be defined. As (3-sqrt(13))/2 is approximately z = 1.5 - (y)/2 where 3<y<4 => z < 1.5 - 1.5 = 0 => z is not in range => a = (3+sqrt(13))/2.

TMAnswered by Teodora M. • Maths tutor

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