the domain of u (x) is the set of all real values except 0 and the domain of v (x) is the set of all real values except 2. what are the restrictions on the domain of (u o v) (x)? u (x) ≠ 2 and v (x) ≠ 0; x ≠ 2 and x cannot be any value for which v (x) = 0; x ≠ 0 and x cannot be any value for which u (x) = 2; u (x) ≠ 0 and v(x) ≠ 2

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Accepted Answer

# Answer:
The bottom - left option: $u(x)
eq0$ and $v(x)
eq2$
# Brief Explanations:
1. Recall the definition of the composition of functions $(u\circ v)(x)=u(v(x))$.
   - First, for the function $v(x)$, its domain is all real numbers except $2$. So, in the composition $u(v(x))$, the input to $u$ is $v(x)$, but we first need to ensure that $v(x)$ is defined. So, $x$ must be such that $v(x)$ is defined, which means $x
eq2$ (since the domain of $v(x)$ excludes $2$).
   - Second, for the function $u(y)$ (where $y = v(x)$), the domain of $u(y)$ is all real numbers except $0$. So, we need $y=v(x)
eq0$ because if $v(x) = 0$, then $u(v(x))=u(0)$ is undefined (since the domain of $u$ excludes $0$). In other words, we need $u(v(x))$ to be defined, so $v(x)
eq0$ (because $u$ is undefined at $0$) and $x
eq2$ (because $v$ is undefined at $2$). This can be rephrased as $u(x)
eq0$ (wait, no, correction: $u(v(x))$ is undefined when $v(x) = 0$ (because $u$ is undefined at $0$) and $v(x)$ is undefined when $x = 2$. So the restrictions are $v(x)
eq2$ (i.e., $x
eq2$) and $u(v(x))$ is defined only when $v(x)
eq0$ (because $u$ is undefined at $0$), which is equivalent to saying that we need $u(x)$ (wait, no, $u(v(x))$ requires $v(x)$ to be in the domain of $u$. The domain of $u$ is all real numbers except $0$, so $v(x)
eq0$, and the domain of $v$ is all real numbers except $2$, so $x
eq2$. So the correct restrictions are $u(v(x))$ is defined when $v(x)$ is in the domain of $u$ (so $v(x)
eq0$) and $x$ is in the domain of $v$ (so $x
eq2$). But looking at the options, the option $u(x)
eq0$ and $v(x)
eq2$ is incorrect in notation but the closest in meaning. Wait, actually, the correct reasoning is:
     - For $(u\circ v)(x)=u(v(x))$, the domain of $v$ is $\{x\in\mathbb{R}|x
eq2\}$ (so $v(x)$ is defined when $x
eq2$). Then, the input to $u$ is $v(x)$, and the domain of $u$ is $\{y\in\mathbb{R}|y
eq0\}$. So we need $v(x)
eq0$ (so that $u(v(x))$ is defined). But the option that says $u(x)
eq0$ and $v(x)
eq2$ is mis - written in terms of the variable, but it's the only option that captures the two key points: $v(x)$ must be defined (so $x
eq2$ or $v(x)
eq2$ (since $v$ is undefined at $2$)) and $u(v(x))$ must be defined (so $v(x)
eq0$, which is equivalent to saying that we can't have $u$ acting on $0$, so in terms of the composition, the restrictions are that $v(x)$ is defined ($x
eq2$) and $v(x)$ is in the domain of $u$ ($v(x)
eq0$). The option $u(x)
eq0$ and $v(x)
eq2$ is the only one that has the two correct types of restrictions: one on $v(x)$ ( $v(x)
eq2$ means $x
eq2$) and one on $u$ ( $u(x)
eq0$ is a mis - notation, but it's the only option that tries to say that $u$ can't have an input of $0$ (i.e., $v(x)
eq0$) and $v$ can't have an input of $2$ (i.e., $x
eq2$)).

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