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a. $f(x)=x^5+\sqrt{x}$

b. $g(y)=\frac{1}{y-2}$

c. $h(z)=\sqrt{z^2-2z-3}$

I know the solutions of all three, but in order to show the proper format of work on paper, what should I do? I'm not sure if my explanations are good enough. Here is my work and solutions for all three:

a. Domain: $x\geqslant0$; Range: $f(x)\geqslant0$.

My reasoning: In order for $f(x)$ to be real, $x$ must be greater or equal to $0$, since only the square root of a non-negative real number gives a real solution. Therefore, the smallest possible $f(x)$ is $0$.

b. Domain: $y\neq2$; Range: $g(y)\neq0$.

My reasoning: A fraction whose denominator equals to $0$ cannot possibly exist. When $y$ is $2$, the denominator is $0$; thus, $y\neq2$. The only value $g(y)$ cannot be is $0$ because the numerator of the function is $1$; the numerator must equal to $0$ in order for $g(y)$ to be $0$. This is not possible, so $g(y)\neq0$.

c. Domain: $z\geqslant3 \cup z\leqslant-1$; Range: $h(z)\geqslant0$.

My reasoning: In order for $h(z)$ to be real, the value inside the square root sign must be greater or equal to $0$, since only the square root of a non-negative real number gives a real solution. Only $z\geqslant3$ and $z\leqslant-1$ make this happen. Therefore, the smallest possible $h(z)$ is $0$.

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    $\begingroup$ Your reasoning appears to be sound. $\endgroup$ – John Wayland Bales Aug 14 '16 at 3:30
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Your explanations all convey the right idea and would likely be more than enough for a high school class. But since you solicited feedback, I'll provide some.

A. You wrote:

$x$ must be greater or equal to $0$, since only the square root of a non-negative real number gives a real solution. Therefore, the smallest possible $f(x)$ is $0$.

This argument is not quite complete: you have not explained the "therefore." You are right that $\sqrt{x}$ is defined only when $x\geq0$ (at least in the setting of real numbers). But you need to explain why this fact implies $x^5+\sqrt{x}\geq0$. Of course, this extra step is very easy to fill in, but as it stands, it is missing from your explanation. Consider, instead, what you would have written if the function had been $g(x)=-x^5+\sqrt{x}$. In this case, you could write the same first sentence that you wrote for $f(x)=x^5+\sqrt{x}$, but it does not follow that $g(x)\geq0$. This shows that you have left something out.

B. You wrote:

A fraction whose denominator equals to $0$ cannot possibly exist.

A mathematician would not speak of "existence" here. What you mean to say is that the expression $(y-2)^{-1}$ is undefined precisely when $y=2$. It is not that no such fraction exists (consider, for example, the fraction $1/0$); it is that such a fraction is meaningless (in this context, at least). A fraction is a string of symbols, not a number. We use (some) fractions to refer to numbers, but the fraction is not the same thing as the number, just as the word cat is not the same thing as a cat. We often get away with confusing a fraction with the number it represents, but strictly speaking, that way of talking doesn't make sense.

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  • $\begingroup$ Awesome, thanks! $\endgroup$ – 关一骏 Aug 14 '16 at 20:27
  • $\begingroup$ Just an FYI: this is summer homework for my upcoming AP Calculus BC course next school semester. $\endgroup$ – 关一骏 Oct 8 '16 at 1:48

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