Let's solve each of these problems step by step.

1. Finding $(f \circ g)(x)$

The composition $(f \circ g)(x)$ means we apply $g(x)$ first and then apply $f$ to the result.

Given: $f(x) = \frac{1}{x}, \quad g(x) = x^2 - 4x$

So: $(f \circ g)(x) = f(g(x)) = f(x^2 - 4x)$

Substitute $g(x) = x^2 - 4x$ into $f(x)$:

$f(x^2 - 4x) = \frac{1}{x^2 - 4x}$

Thus: $(f \circ g)(x) = \frac{1}{x^2 - 4x}$

Domain of $(f \circ g)(x)$

The domain of $(f \circ g)(x)$ is the set of all $x$ such that the expression inside $f(x)$ is defined. Since $f(x) = \frac{1}{x}$, the expression $x^2 - 4x \neq 0$.

Solve for when $x^2 - 4x = 0$:

$x(x - 4) = 0 \implies x = 0 \text{ or } x = 4$

So, the domain of $(f \circ g)(x)$ is all real numbers except $0$ and $4$.

Domain in interval notation: $(-\infty, 0) \cup (0, 4) \cup (4, \infty)$

2. Finding $(g \circ f)(x)$

The composition $(g \circ f)(x)$ means we apply $f(x)$ first and then apply $g$ to the result.

$(g \circ f)(x) = g(f(x)) = g\left(\frac{1}{x}\right)$

Substitute $f(x) = \frac{1}{x}$ into $g(x)$:

$g\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^2 - 4 \left(\frac{1}{x}\right)$

Simplify:

$= \frac{1}{x^2} - \frac{4}{x}$

Thus: $(g \circ f)(x) = \frac{1}{x^2} - \frac{4}{x}$

Domain of $(g \circ f)(x)$

The domain of $(g \circ f)(x)$ is the set of all $x$ such that the expression inside $g(x)$ is defined. Since $f(x) = \frac{1}{x}$, we need $x \neq 0$. Additionally, there are no further restrictions for the result of $g\left(\frac{1}{x}\right)$, so the only restriction is $x \neq 0$.

Domain in interval notation: $(-\infty, 0) \cup (0, \infty)$

3. Finding $(f \circ f)(x)$

The composition $(f \circ f)(x)$ means we apply $f(x)$ first and then apply $f$ again to the result.

$(f \circ f)(x) = f(f(x)) = f\left(\frac{1}{x}\right)$

Substitute $f(x) = \frac{1}{x}$ into itself:

$f\left(\frac{1}{x}\right) = \frac{1}{\frac{1}{x}}$

Simplify:

$= x$

Thus: $(f \circ f)(x) = x$

Domain of $(f \circ f)(x)$

The domain of $(f \circ f)(x)$ is the domain of $f(x)$ applied twice. Since $f(x) = \frac{1}{x}$, $x \neq 0$. Therefore, the domain of $(f \circ f)(x)$ is all real numbers except $0$.

Domain in interval notation: $(-\infty, 0) \cup (0, \infty)$

Summary of Results:

1. $(f \circ g)(x) = \frac{1}{x^2 - 4x}$

   • Domain: $(- \infty, 0) \cup (0, 4) \cup (4, \infty)$

2. $(g \circ f)(x) = \frac{1}{x^2} - \frac{4}{x}$

   • Domain: $(- \infty, 0) \cup (0, \infty)$

3. $(f \circ f)(x) = x$

   • Domain: $(- \infty, 0) \cup (0, \infty)$

Would you like more details or have any questions?

Related Questions:

1. What is the general process for finding the domain of a composition of functions?
2. How do you determine the range of a composition of functions?
3. Can you provide an example of a composition of functions where the domain is restricted by both functions?
4. What are the properties of function composition?
5. How would the composition change if one of the functions were not defined for all real numbers?

Tip:

When finding the domain of a composition, always ensure that the input of the inner function falls within its domain, and the output must also be suitable for the outer function.