I know what domain and range are, at least I think I do, but recently I took a test in which they asked me to determine the domain and range of a function I remember only very faintly now. In the test, I answered what seemed logical to me, however, later on I found that the answer was completely wrong when I looked at a graph of the function. My question is: is there a way to determine the domain and range of a function without looking at its graph?
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- $\begingroup$ have you got an example? $\endgroup$Dr. Sonnhard Graubner– Dr. Sonnhard Graubner2015-11-08 02:42:18 +00:00Commented Nov 8, 2015 at 2:42
- 1$\begingroup$ I would like to know how to do it generally, but if an example would help clarify the process I would say $\ln (|\frac {x}{1+x^2}|)$ $\endgroup$GuPe– GuPe2015-11-08 03:01:01 +00:00Commented Nov 8, 2015 at 3:01
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1 Answer
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2 Unless explicitly stated, the domain of a function $f(x)$ is all possible values of $x$ for which $f(x)$ is defined. For example the function $$f(x)=\frac{1}{x}$$
is defined for $$\{x\in\mathbb{R}\,|\,x\neq 0\}$$
The range of a function can be determined by finding the absolute maxima or minima. In this case $f(x)$ has no absolute maxima or minima, so the range of $f(x)$ is the entire real number line $\mathbb{R}$. For another function
$$g(x)=x^2-4$$
note that the first derivative is
$$g^\prime(x)=2x$$
which equals $0$ at $x=0$. Thus this is a minimum or maximum. The second derivative at this point is
\begin{gather} g^{\prime\prime}(x)=2\\ g^{\prime\prime}(0)=2 \end{gather}
Because the second derivative is positive, $x=0$ is a local minimum. Note that $g(x)$ at this point is equal to $g(0)=-4$. Because the limits \begin{gather} \lim_{x\to\infty}x^2-4=\infty\\ \lim_{x\to -\infty}x^2-4=\infty \end{gather} are both $>-4$, no value of $g(x)$ can be lower than $-4$. Thus it is an absolute maximum and the range of $g(x)$ is $$\{x\in\mathbb{R}\,|\,-4\leq x\}$$
- $\begingroup$ What would happen with ceiling and floor functions? $\endgroup$GuPe– GuPe2015-11-08 03:22:46 +00:00Commented Nov 8, 2015 at 3:22
- $\begingroup$ The same process would still apply, except you would have to have $x\in\mathbb{N}$. Visualizing the graph of the function helps immensely. $\endgroup$cheesyfluff– cheesyfluff2015-11-08 03:26:11 +00:00Commented Nov 8, 2015 at 3:26