Interpretting Functions: Relating Domain to Context and Graphs (CCSS.F-IF.5)

The denominator $x-3$ is zero at $x=3$, so $x=3$ must be excluded. Therefore, the domain is all real numbers except $3$. Choice A is wrong because it excludes $-3$ instead and still allows $x=3$ (which makes the denominator zero). Choice C incorrectly restricts to integers only. Choice D adds an irrelevant condition ($x>100$) not implied by the function.

Require $5-2x\ge 0\Rightarrow -2x\ge -5\Rightarrow x\le \tfrac{5}{2}$, so the domain is $(-\infty,\tfrac{5}{2}]$. Choice A confuses domain with a typical range for square roots ($y\ge 0$). Choice B restricts to integers only though any real $x\le \tfrac{5}{2}$ is allowed. Choice C includes $x>\tfrac{5}{2}$, which makes the radicand negative.

In this context $x$ counts tickets, so $x$ must be a nonnegative integer: $\{0,1,2,3,\ldots\}$. Choice B incorrectly allows all real numbers $\ge 0$, including non-integers. Choice C allows negative and non-integer values. Choice D imposes an irrelevant condition ($x>100$) not given by the situation.

We need $x-1>0$ so the square root is defined and nonzero in the denominator; thus $x>1$ and the domain is $(1,\infty)$. Choice A incorrectly includes $x=1$, which makes the denominator $0$. Choice B restricts to integers only even though any real $x>1$ works. Choice D includes values $x\le 1$, where the expression is undefined.

We need $3-x>0\Rightarrow x<3$, so the domain is $(-\infty,3)$. Choice B includes $x\ge 3$, where the logarithm is undefined. Choice C is a typical range-like set and not the correct domain here. Choice D restricts to integers only and also includes values $\ge 3$ that are not allowed.

The denominator must be nonzero: $x+3\ne 0\Rightarrow x\ne -3$. So the domain is $(-\infty,-3)\cup(-3,\infty)$. Choice B confuses the domain with the range (here $y\ne 2$). Choice C wrongly restricts to integers only. Choice D incorrectly includes $x=-3$, which makes the denominator zero.

The radicand must be nonnegative: $8-2x\ge 0\Rightarrow x\le 4$. Thus the domain is $(-\infty,4]$. Choice A is the range of a square-root expression (confusing range with domain). Choice B wrongly limits to integers only. Choice D includes values $x>4$ that make $8-2x<0$, which is not allowed.

Restrictions: (1) Square root requires $x-2\ge 0\Rightarrow x\ge 2$. (2) Denominator nonzero: $x\ne 0$ and $x\ne 5$. Since $x\ge 2$, $x=0$ is irrelevant, but $x=5$ must be excluded. So the domain is $[2,5)\cup(5,\infty)$. Choice A confuses the domain with a typical range for square roots. Choice B wrongly includes $x=5$. Choice C restricts to integers only and also includes the invalid $x=5$.

In context, $x$ represents hours and is limited by the shop hours: $0\le x\le 8$. So the domain is $[0,8]$. Choice A confuses domain with range: when $0\le x\le 8$, the cost ranges from $60$ to $180$. Choice C incorrectly restricts to integers only (hours could be fractional). Choice D is an irrelevant condition not implied by the situation.

For a set of ordered pairs, the domain is the set of input values (first coordinates): $\{-3,0,2,5\}$. Choice B is the range (outputs), confusing range with domain. Choice C wrongly claims all integers. Choice D is an irrelevant condition.