Question

1. let ( f(x) = \frac{3}{x - 4} ) and ( g(x) = x + 2 ). find ( (g circ f)(x) ) and state its domain.

\\( 2x - 1 \\); the domain of ( (g circ f)(x) ) is the set of all real numbers except 9
\\( x + 4 \\); the domain of ( (g circ f)(x) ) is the set of all real numbers except 10
\\( \frac{2x + 2}{x - 3} \\); the domain of ( (g circ f)(x) ) is the set of all real numbers except 2
\\( \frac{2x - 5}{x - 4} \\); the domain of ( (g circ f)(x) ) is the set of all real numbers except 4

Explanation:

Step 1: Find the composition $(g\circ f)(x)$

By the definition of function composition, $(g\circ f)(x)=g(f(x))$. Given $f(x)=\frac{3}{x - 4}$ and $g(x)=x + 2$, we substitute $f(x)$ into $g(x)$:

$$
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$$

To combine these terms, we get a common denominator:

$$
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$$

Step 2: Determine the domain of $(g\circ f)(x)$

The domain of a composition $(g\circ f)(x)$ must exclude values that make $f(x)$ undefined and values that make $g(f(x))$ undefined.

• For $f(x)=\frac{3}{x - 4}$, the denominator cannot be zero, so $x

eq4$.

• For $g(f(x))=\frac{2x - 5}{x - 4}$, the denominator is also zero when $x = 4$. There are no other restrictions from $g(x)$ (since $g(x)=x + 2$ is a polynomial and defined for all real numbers).

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

Answer:

The correct option is the last one: $\boldsymbol{\frac{2x - 5}{x - 4}}$; the domain of $(g\circ f)(x)$ is the set of all real numbers except 4.