f(x)=x2-5x+4

g(x)=x-1

Find (fg)(x)

(fg)(x)=

The domain of (fg)(x) is x≠

To find \((fg)(x)\), we multiply the functions \(f(x)\) and \(g(x)\):

\[
f(x) = x^2 - 5x + 4
\]
\[
g(x) = x - 1
\]

### Step 1: Multiply \(f(x)\) and \(g(x)\)
\[
(fg)(x) = f(x) \cdot g(x)
\]
\[
(fg)(x) = (x^2 - 5x + 4)(x - 1)
\]

Now, use the distributive property to multiply:

\[
(fg)(x) = x^2(x - 1) - 5x(x - 1) + 4(x - 1)
\]
\[
(fg)(x) = x^3 - x^2 - 5x^2 + 5x + 4x - 4
\]
\[
(fg)(x) = x^3 - 6x^2 + 9x - 4
\]

Thus, 
\[
(fg)(x) = x^3 - 6x^2 + 9x - 4
\]

### Step 2: Find the domain of \((fg)(x)\)
The domain of a product of functions is the set of all values for which both \(f(x)\) and \(g(x)\) are defined. Since both \(f(x)\) and \(g(x)\) are polynomials, their domain is all real numbers, except where \(g(x) = 0\), because division by zero would be undefined.

\[
g(x) = x - 1
\]
Setting \(g(x) = 0\), we get:
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]

Therefore, the domain of \((fg)(x)\) is all real numbers except \(x = 1\).

### Final Answer:
\[
(fg)(x) = x^3 - 6x^2 + 9x - 4
\]
The domain of \((fg)(x)\) is \(x \neq 1\).

Would you like more details or explanations?

Here are some related questions for you to consider:
1. How do you factorize \( (fg)(x) = x^3 - 6x^2 + 9x - 4 \)?
2. What is the range of \( (fg)(x) \)?
3. How do you graph \( (fg)(x) = x^3 - 6x^2 + 9x - 4 \)?
4. Can you find the x-intercepts of \( (fg)(x) \)?
5. How do you find the critical points of \( (fg)(x) \)?

**Tip:** To check the domain of composite or product functions, always check where each function is undefined first.

f(x) = x^2 - 5x + 4, g(x) = x - 1, Find (fg)(x) and the domain of (fg)(x)

Learn how to find the product of two functions, f(x) = x^2 - 5x + 4 and g(x) = x - 1, and determine the domain of (fg)(x). This step-by-step algebraic solution explores polynomial multiplication and domain restrictions. Suitable for Grades 9-12.

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