Answer to Question #142149 in Calculus for chan

Question

1.Let f(x)=x+1,g(x)=x^2-3. Solve the following equation for x [(fog)(x)+(gof)(x)=0]
2.Determine whether the following function is invertible or not, if it is , find its inverse,
g(x)=x-1/3x+2

Answer 1

1. $fog(x) =f(g(x)) =(x^2-3)+1=x^2-2, \\ gof(x) =g(f(x)) =(x+1)^2-3=x^2+2x-2.$

So

$fog(x) +gof(x) =2x^2+2x-4=0. \\ x^2+x-2=0, \\ x_1=-2, x_2=1.$

Therefore, there are two roots of this equation.

2.$\;g(x) =x-\dfrac{1}{3x}+2$

The invertible function should be strictly monotonic, so it cannot have the same value for different x. So it should have no more than 1 root. Let us determine the roots of g(x)

$0=x-\dfrac{1}{3x}+2, \\ 0=3x^2+6x-1. \\ D=b^2-4ac =36+12=48>0$

Therefore, the function has two roots and it can't be monotonic and invertible.