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  1. Find the domain and range of the functions:

a. $y=\sqrt{x+3}$

domain: $[-3, \infty)$ and range: $[0,\infty)$

b. $y=\frac{1}{x+4}$

domain:$(-4,\infty)$ and range: $(0,\infty)$

c. $y=\frac{\sqrt{x}}{x-3}$

domain:$[0,3) \cup(3,\infty)$ and range: $[0,\infty]$

Can anyone confirm my answers, and if any of the answers are wrong, provide an explanation.

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  • $\begingroup$ negative numbers below -4 will do in part b for the domain, and the range will also change accordingly, and range is wrong for part c also, plug in $x=2$ and see $\endgroup$ – Nikunj Feb 27 '16 at 22:16
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a) Your answer is correct.

b) The domain is incorrect. $x \rightarrow x+4$ is a bijection and every non-zero real has a multiplicative inverse. The denominator is $0$ at $x=-4$, so the domain is $(-4, \infty) \cup (-\infty, -4)$. The range is also incorrect; it is $(0, \infty) \cup (-\infty, 0)$. This is clear by considering the inverse $\frac {1-4x}{x}$.

c) The domain is fine. The range is incorrect. This interesting function is, in fact, surjective onto $\mathbb{R}$. It vanishes at $x=0$ and has a vertical asymptote at $x=3$, approaching $\infty$ from the right and $-\infty$ from the left. As $x \rightarrow \infty$ it is positive and becomes arbitrarily small (that is, it tends to $0$).

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  • $\begingroup$ the range for c would be all real numbers then? $\endgroup$ – kero Feb 27 '16 at 23:35
  • $\begingroup$ Yes. That is what "surjective onto $\mathbb{R}$" means. $\endgroup$ – MathematicsStudent1122 Feb 27 '16 at 23:36

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