Match the corresponding function formula with each function when h(x) = 3x + 2 and g(x) = 2 x . 1. k(x) = 2x + 3x + 2                    k(x) = g(x) ∘ h(x) 2. k(x) = 2x - 3x - 2                      k(x) = g(x) + h(x) 3. k(x) = (3x)2-x + 2-x + 1            k(x) = h(x) ÷ g(x) 4. k(x) = 23x + 2                              k(x) = g(x) - h(x) 5. k(x) = 2x(3x + 2)                      k(x) = g(x) × h(x) 6. k(x) = 3(2x) + 2                  k(x) = h(x) ∘ g(x)

The correct functions are
h(x) = 3x + 2  
g(x) = 2^x

I proceed to calculate the different values of K (x)
case 1) k(x) = g(x) ∘ h(x)
g(x) ∘ h(x)=g(h(x))
k(x) = [2]^[3x + 2]> is the option 4.k(x) = 2^(3x + 2 )
 
case 2) k(x) = g(x) + h(x)
k(x) = [2^x]+[3x + 2]=2^x+3x+2> is the option 1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)
k(x) = [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)
is the option 3.)  k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)
k(x) = [2^x]-[3x + 2]=2^x-3x-2> is the option 2.) k(x) = 2^x - 3x - 2 

case 5) k(x) = g(x) × h(x)
k(x) = [2^x]*[3x + 2]> is the option 5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)
h(x) ∘ g(x)=h(g(x))
k(x)=3*[2^x]+2> is the option 6.k(x) = 3(2^x) + 2 

Two because two of one thing is two.