Consider the function f(x), g(x), h(x) as given below. Show that (fog)oh = fo(goh)

f(x) = x square, g(x) = 2x and h(x) = x + 4


To prove that \((f \circ g) \circ h = f \circ (g \circ h)\), we'll use the given functions:

- \( f(x) = x^2 \)
- \( g(x) = 2x \)
- \( h(x) = x + 4 \)

### Step 1: Calculate \((f \circ g)(x)\)
This is the composition of \(f(x)\) and \(g(x)\). It means applying \(g(x)\) first and then applying \(f(x)\) to the result.

\[
(f \circ g)(x) = f(g(x))
\]

Substitute \(g(x)\) into \(f(x)\):

\[
f(g(x)) = f(2x) = (2x)^2 = 4x^2
\]

So, \((f \circ g)(x) = 4x^2\).

### Step 2: Calculate \(((f \circ g) \circ h)(x)\)
Now, we apply \(h(x)\) first and then apply \((f \circ g)\) to the result.

\[
((f \circ g) \circ h)(x) = (f \circ g)(h(x))
\]

Substitute \(h(x)\) into \((f \circ g)(x)\):

\[
(f \circ g)(h(x)) = 4(h(x))^2
\]

Substitute \(h(x) = x + 4\):

\[
(f \circ g)(h(x)) = 4(x + 4)^2
\]

Expand the square:

\[
4(x + 4)^2 = 4(x^2 + 8x + 16) = 4x^2 + 32x + 64
\]

So, \(((f \circ g) \circ h)(x) = 4x^2 + 32x + 64\).

### Step 3: Calculate \((g \circ h)(x)\)
This is the composition of \(g(x)\) and \(h(x)\). It means applying \(h(x)\) first and then applying \(g(x)\) to the result.

\[
(g \circ h)(x) = g(h(x))
\]

Substitute \(h(x)\) into \(g(x)\):

\[
g(h(x)) = 2(h(x)) = 2(x + 4)
\]

Distribute:

\[
g(h(x)) = 2x + 8
\]

So, \((g \circ h)(x) = 2x + 8\).

### Step 4: Calculate \((f \circ (g \circ h))(x)\)
Now, apply \(f(x)\) to the result of \((g \circ h)(x)\).

\[
(f \circ (g \circ h))(x) = f(g(h(x)))
\]

Substitute \(g(h(x)) = 2x + 8\) into \(f(x)\):

\[
f(g(h(x))) = (2x + 8)^2
\]

Expand the square:

\[
(2x + 8)^2 = 4x^2 + 32x + 64
\]

So, \((f \circ (g \circ h))(x) = 4x^2 + 32x + 64\).

### Conclusion
Since:

\[
((f \circ g) \circ h)(x) = 4x^2 + 32x + 64
\]

and

\[
(f \circ (g \circ h))(x) = 4x^2 + 32x + 64
\]

We have shown that:

\[
(f \circ g) \circ h = f \circ (g \circ h)
\]

**Do you want more details or have any questions?**

Here are some related questions you might find interesting:

1. How do you generalize the associative property of function composition for arbitrary functions?
2. What is the difference between left and right composition in functions?
3. How does function composition relate to real-world applications, such as in computer science?
4. What are the conditions under which function composition is not associative?
5. How do you interpret the composition of functions graphically?

**Tip:** When dealing with function compositions, always remember to apply functions from the innermost to the outermost.

Consider the function f(x), g(x), h(x) as given below. Show that (f∘g)∘h = f∘(g∘h). f(x) = x^2, g(x) = 2x, and h(x) = x + 4.

Explore the proof of the associative property of function composition with the functions f(x) = x^2, g(x) = 2x, and h(x) = x + 4. This detailed guide walks through the step-by-step solution, ideal for students in Grades 10-12 learning advanced algebra concepts.

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