Precalculus ! 1. find (fog)(x) when f(x)=2/x+3 and g(x)=1/2x. 2. find (f*g)(x) when f(x)=(sqrt)x+3/x and g(x)=(sqrt)x+3/2x. 3. find (f+g)(x) when f(x)=x^3-2x^2+1 and g(x)=4x^3-5x+7

1)  

2)

3)    

Step-by-step explanation:

1) Given : and

To find :            

Solution :  

We know that,

i.e g(x) within f(x),

Therefore,  

2) Given : and

To find :            

Solution :

We know that,

Therefore,  

3) Given : and

To find :            

Solution :

We know that,

Therefore,

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

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 + > 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+> 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-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

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