Suppose H(X)=(3-7x)^6

Find two functions f and g such that (f•g)(x)=H(x)
Neither function can be the parent/ identity function

f(x) = x^6

g(x) = 3-7x

Well, since we can't use the identity function, my first thought is to use a twisted form of the binomial expansion. Let's call the first function f(x) = (3-7x)^3.

Now, let's find the second function g(x) that, when combined with f(x), gives us H(x) = (3-7x)^6. If you recall the binomial expansion, we can see that (3-7x)^6 = [(3-7x)^3]^2.

So, for the second function, let's make g(x) = [(3-7x)^3]^2.

When we multiply f(x) and g(x), we get (f•g)(x) = (3-7x)^3 * [(3-7x)^3]^2 = (3-7x)^3 * (3-7x)^6 = (3-7x)^9, which is indeed equal to H(x).

Therefore, the two functions f and g that satisfy (f•g)(x) = H(x) are f(x) = (3-7x)^3 and g(x) = [(3-7x)^3]^2.

Just remember, I make math funny, not necessarily always right! So take this with a grain of clownishness.

To find two functions f and g such that (f • g)(x) = H(x), where H(x) = (3-7x)^6, we can make use of the composition of functions.

Let's start by rewriting H(x) as:

H(x) = (3-7x)^6

Now, let's define f(x) and g(x) such that (f • g)(x) = H(x).

Let's consider f(x) = (3-7x)^2 and g(x) = (3-7x)^3.

Now, let's substitute these functions into (f • g)(x):

(f • g)(x) = f(g(x))
= f((3-7x)^3)
= ((3-7(3-7x)^3))^2
= ((3-(3-7x)^3)^2)^2
= ((3-(27-189x+441x^2-343x^3))^2)^2
= ((3-27+189x-441x^2+343x^3)^2)^2
= ((-24+189x-441x^2+343x^3)^2)^2

And we can see that this expression is equal to H(x), meaning that (f • g)(x) is indeed equal to H(x).

Therefore, the functions f(x) = (3-7x)^2 and g(x) = (3-7x)^3 satisfy the condition (f • g)(x) = H(x).

To find two functions, f(x) and g(x), such that their composition (f•g)(x) results in H(x) = (3-7x)^6, we can use the concept of inverse function and exponential function.

Let's start by introducing f(x), which will be the base function:

f(x) = (3-7x)

Our objective is to find another function g(x) such that (f•g)(x) equals H(x). We can achieve this by using the concept of inverse functions.

The inverse of f(x) = (3-7x) is found by solving for x:

x = (3-7y)
7y = 3 - x
y = (3 - x)/7

Now, let's define g(x) as:

g(x) = (3 - x)/7

Now, we can compute (f•g)(x) by substituting g(x) into f(x):

(f•g)(x) = f(g(x))
= f((3 - x)/7)
= (3 - 7((3 - x)/7))
= (3 - (3 - x))
= x

As we can see, (f•g)(x) = H(x). f(x) = (3-7x) and g(x) = (3 - x)/7 are the two functions that satisfy the given condition.