Composition of Functions

If f(x) = 2x+1
find

A. f o f(x)
f(f(x)) = f(2x+1)
f(2x+1) = 2(2x+1)
f(f(x)) = 4x +2

B. f o f o f o f(x)

^ I don't get how to solve this one

C. f o f o f o f o f ...o f(x) n functions of f

^ I also don't get how to solve this one

given f(x) = ax+b and g(x) = cx+d and
f o g(x) = g o f(x), what is the relationship between a,b,c,d?

f(x) = 2x+1

f(f) = 2f+1 = 2(2x+1)+1 = 4x+3
f(f(f)) = 2(4x+3)+1 = 8x + 7
f(f(f(f(x)))) = 2(8x+7)+1 = 16x + 15

fn(x) = 2^n * (x+1) - 1

f(g) = ag+b = a(cx+d)
g(f) = cf+d = c(ax+b)
so, if f(g) = g(f)

acx + ad = acx + bc
ad = bc

A. To find f o f(x), we need to substitute the function f(x) into itself repeatedly. Let's start by finding f(f(x)):

f(f(x)) = f(2x+1)

Now substitute f(x) back into the equation:

f(f(x)) = f(2x+1) = 2(2x+1) + 1 = 4x + 2 + 1 = 4x + 3

So, f o f(x) is equal to 4x + 3.

B. To find f o f o f o f(x), we need to substitute the function f(x) into itself repeatedly. This is equivalent to applying the function multiple times. Therefore:

f o f o f o f(x) = f(f(f(f(f(x)))))

We can simplify this to f^5(x), where f^5 represents applying the function f(x) five times. So, it would be:

f o f o f o f(x) = f^5(x)

C. Similarly, for f o f o f o f o f ...o f(x) n functions of f, we can represent this as f^n(x), where f^n represents applying the function f(x) n times.

Given f(x) = ax + b and g(x) = cx + d, if f o g(x) = g o f(x), it means applying f after g and applying g after f yield the same result. Mathematically, we can represent this as:

f(g(x)) = g(f(x))

Let's substitute the functions f(x) and g(x) into the equation:

f(g(x)) = f(cx + d) = a(cx + d) + b = acx + ad + b

g(f(x)) = g(ax + b) = c(ax + b) + d = acx + cb + d

Since f(g(x)) = g(f(x)), this means acx + ad + b = acx + cb + d. By equating the coefficients of x and the constants, we can see that:

ad + b = cb + d
ad - cb = d - b

Therefore, the relationship between a, b, c, and d is that ad - cb = d - b.

To solve part B, you need to apply the function f(x) repeatedly. Each time you apply the function, you replace x with the expression 2x+1. So, we have:

f(f(f(f(x)))) = f(f(f(2x+1)))

To simplify, we can start by expanding the innermost function:

f(2x+1) = 2(2x+1) + 1 = 4x + 2 + 1 = 4x + 3

Now we can substitute this result back into the expression:

f(f(f(2x+1))) = f(f(4x+3))

Expanding the next function:

f(4x+3) = 2(4x+3) + 1 = 8x + 6 + 1 = 8x + 7

Substituting again:

f(f(4x+3)) = f(8x+7)

And one more time:

f(8x+7) = 2(8x+7) + 1 = 16x + 14 + 1 = 16x + 15

So, the final result is:

f(f(f(f(x)))) = 16x + 15

Now let's move on to part C. The question asks about the composition of the function f(x) with itself n times. In other words, we want to find f(f(f(...f(x))...)), where the function f is applied n times.

We can use the same approach as before, applying f(x) repeatedly:

f(f(f(...f(x))...)) = f(f(f(...f(ax+b))...))

Let's denote this composition as C_n(x), where n represents the number of times the function f is applied.

We know that:

C_1(x) = f(ax+b) = a(ax+b) + b = a^2x + ab + b

Applying the function f one more time:

C_2(x) = f(a^2x + ab + b) = a(a^2x + ab + b) + b = a^3x + a^2b + ab + b

We can see that the pattern is a^nx + (a^(n-1)b + a^(n-2)b + ... + ab + b)

So, the expression for C_n(x) is:

C_n(x) = a^nx + (a^(n-1)b + a^(n-2)b + ... + ab + b)

Moving on to the last question, let's consider the composition of functions f o g(x) = g o f(x).

First, we need to evaluate f o g(x):

f(g(x)) = f(cx+d) = a(cx+d) + b = acx + ad + b

Now, let's evaluate g o f(x):

g(f(x)) = g(ax+b) = c(ax+b) + d = acx + cb + d

Since we are given that f o g(x) = g o f(x), we can equate the expressions:

acx + ad + b = acx + cb + d

This equation implies that ad + b = cb + d.

So, the relationship between a, b, c, and d is that ad + b = cb + d.

To solve B. and C., you need to repeatedly apply the function f(x) multiple times.

B. To find f o f o f o f(x), you need to apply the function f(x) four times. Start by applying f(x) to x, then applying f(x) to the result, and so on, four times. It can be represented as:

f o f o f o f(x) = f(f(f(f(x))))

So, substitute f(x) into the equation four times:

f(f(f(f(x)))) = f(f(f(2x + 1)))
= f(f(2(2x + 1) + 1))
= f(f(4x + 3))
= f(2(4x + 3) + 1)
= f(8x + 7)
= 2(8x + 7) + 1
= 16x + 15

Therefore, f o f o f o f(x) = 16x + 15.

C. Finding the composition of f(x) applied n times, represented as f o f o f o f o f ...o f(x) n functions of f, can be done similarly. You need to apply the function f(x) n times.

f o f o f o f o f ...o f(x) n functions of f = f(f(f(f(f(...(f(x))...))))

So, substitute f(x) into the equation n times:

f(f(f(f(f(...(f(x))...)))) = f(f(f(f(f(...(2x + 1)...)))))
= f(f(f(f(...(2(2x + 1) + 1)...)))))
= f(f(f(...(2(2(2x + 1) + 1) + 1)...)))))
= f(f(...(2(2(2(2x + 1) + 1) + 1) + 1)...)))))
= f(...(2(2(2(2(2x + 1) + 1) + 1) + 1) + 1)...)))))
= (...(2^(n-1)(2x + 1) + 1)...)

Therefore, f o f o f o f o f ...o f(x) n functions of f = 2^(n-1)(2x + 1) + 1.

Now, for the last question:

Given that f(x) = ax + b and g(x) = cx + d, if f o g(x) = g o f(x), we can equate the two compositions:

f(g(x)) = g(f(x))

Now, let's substitute the given expressions for f(x) and g(x):

a(cx + d) + b = c(ax + b) + d

Simplifying this equation:

acx + ad + b = acx + cb + d

The terms on both sides have the same coefficients for x, so the equation simplifies further:

ad + b = cb + d

To find the relationship between a, b, c, and d, we can rearrange the equation:

ad - cb = d - b

Hence, the relationship between a, b, c, and d is that the difference between ad and cb is equal to the difference between d and b.