# Chapter 1 - Section 1.7 - Combinations of Functions; Composite Functions - Exercise Set: 16

$(-\infty, 2) \cup (2, \frac{10}{3}) \cup (\frac{10}{3}, +\infty)$.

#### Work Step by Step

The denominator is not allowed to be equal to zero as division of zero leads to an undefined expression. Look for the real numbers that will make the denominators of the given function equal to zero. The expression $\dfrac{4}{x-2}$ is undefined when $x=2$. To find the value of $x$ that will make the denominator $\dfrac{4}{x-2} -3$ equal to $0$, equate the denominator to zero then solve the equation to obtain: $\dfrac{4}{x-2} - 3 = 0 \\\dfrac{4}{x-2} = 3$ Cross-multiply: $\\4 = 3(x-2) \\4 = 3x - 6 \\4+6 = 3x \\10 = 3x \\\frac{10}{3} = x$ Thus, the value of $x$ can be any real number except $2$ and $\frac{10}{3}$. Therefore, the domain is $(-\infty, 2) \cup (2, \frac{10}{3}) \cup (\frac{10}{3}, +\infty)$.

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