Free worksheet on functions with composites and inverses, including solutions.

The image you provided is a worksheet titled "FUNCTIONS: EQUATIONS WITH COMPOSITES & INVERSES." It contains several problems involving composite functions and inverses. Below, I will solve one of the problems as an example and explain the solution step by step.

Problem to Solve:

C3:
- \( f(x) = x - 1 \)
- \( g(x) = x^2 \)
- Solve \( ff(x) = g^{-1}(x) \)

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Step-by-Step Solution:

#### Step 1: Understand the problem
We need to solve the equation \( ff(x) = g^{-1}(x) \). This involves:
1. Finding \( ff(x) \) (the composition of \( f \) with itself).
2. Finding \( g^{-1}(x) \) (the inverse of \( g(x) \)).
3. Setting \( ff(x) \) equal to \( g^{-1}(x) \) and solving for \( x \).

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#### Step 2: Compute \( ff(x) \)
The function \( f(x) = x - 1 \). To find \( ff(x) \), we compute \( f(f(x)) \):
\[
f(f(x)) = f(x - 1)
\]
Substitute \( x - 1 \) into \( f(x) \):
\[
f(x - 1) = (x - 1) - 1 = x - 2
\]
Thus,
\[
ff(x) = x - 2
\]

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#### Step 3: Find \( g^{-1}(x) \)
The function \( g(x) = x^2 \). To find the inverse \( g^{-1}(x) \), we solve for \( x \) in terms of \( y \) where \( y = g(x) \):
\[
y = x^2
\]
Taking the square root of both sides:
\[
x = \pm \sqrt{y}
\]
Since \( g(x) = x^2 \) is not one-to-one over all real numbers, we need to specify the domain. Assuming the principal (non-negative) square root (common in such problems unless otherwise stated):
\[
g^{-1}(x) = \sqrt{x}
\]

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#### Step 4: Set \( ff(x) \) equal to \( g^{-1}(x) \)
We now set \( ff(x) = g^{-1}(x) \):
\[
x - 2 = \sqrt{x}
\]

#### Step 5: Solve the equation \( x - 2 = \sqrt{x} \)
To eliminate the square root, square both sides:
\[
(x - 2)^2 = (\sqrt{x})^2
\]
\[
(x - 2)^2 = x
\]
Expand the left side:
\[
x^2 - 4x + 4 = x
\]
Rearrange the equation to form a standard quadratic equation:
\[
x^2 - 4x + 4 - x = 0
\]
\[
x^2 - 5x + 4 = 0
\]

#### Step 6: Solve the quadratic equation
Factorize the quadratic equation:
\[
x^2 - 5x + 4 = (x - 4)(x - 1) = 0
\]
Set each factor to zero:
\[
x - 4 = 0 \quad \text{or} \quad x - 1 = 0
\]
\[
x = 4 \quad \text{or} \quad x = 1
\]

#### Step 7: Verify the solutions
We need to check if both solutions satisfy the original equation \( x - 2 = \sqrt{x} \).

1. For \( x = 4 \):
\[
4 - 2 = \sqrt{4}
\]
\[
2 = 2 \quad \text{(True)}
\]

2. For \( x = 1 \):
\[
1 - 2 = \sqrt{1}
\]
\[
-1 = 1 \quad \text{(False)}
\]

Thus, \( x = 1 \) is not a valid solution because it does not satisfy the original equation.

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Final Answer:

The only valid solution is:
\[
\boxed{4}
\]