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Well, I do understand what Df is and how you find it in simple equations, however, I am kinda confused in "complex" functions.

For example, the following functions: 1* f(x)=x^3+x^2-x-1 , Df=R (however, I don't understand why.)

2* f(x)=2x/(1+x^2), we have to find Df for 1+x^2, therefore Df=R, I do understand this one.

3* f(x)=2x^2-x^4 , Df=R (Don't know why)

4* f(x)=x^2/(x-2), Df=R (don't know why)

Could anyone explain why is the Df always R in this cases?

user2041143
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1 Answers1

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I'm skeptic about your first sentence.

$D_f$ is the domain of $f$, and the exercises ask you to find the maximal subset of $\Bbb R$ such that the given $f$ is defined on each point of that subset.

In the given examples, the only problem of being not defined can arise by the denominators, and you only have to know that $\frac ab$ is defined iff $b\ne 0$. (In other domain questions, you also have to use like '$\sqrt u$ is defined iff $u\ge 0$' and '$\log u$ iff $u>0$', etc.

So now, 1. $f(x)=x^3+x^2-x-1$ have no problem, it is defined for all $x\in\Bbb R$. The same holds for 3.

For 2., we have that $1+x^2\ge 1$ for all real $x$, so the denominator can't be $0$, that's why $D_f=\Bbb R$ again.

But 4. is not defined for $x=2$, so $D_f=\Bbb R\setminus\{2\}$.

Berci
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  • I understand 4th one, however first and third are still confusing me, why x∈R? How do we know that? Is there anything I need to know ? Or any function with degree of 3,2 is always in R? – user2041143 Mar 03 '13 at 11:43
  • Consider the machine that outputs $r^3$ whenever you input the number $r$. For which real numbers will it give an output? For which inputs $r$ will it get frozen and don't output anything? Then consider a similar machine for $r\mapsto 1/r$. – Berci Mar 03 '13 at 11:45
  • It will run for a "long" time, probably. Do you mean that r^3 doesn't have an end, therefore it's in R? What happens with other elements of the functions then? – user2041143 Mar 03 '13 at 11:46
  • I simply want to mean that the function $x\mapsto x^3$ is defined for all real numbers $x$: any real number can be plugged in place of $x$, and the resulting expression will indeed give one well-determined real number. Same holds for the functions in 1. and 3. – Berci Mar 03 '13 at 11:48
  • I understand. I remember we studied that a few weeks ago. However, do we check the other elements such as x^2 in the function or we determinate the Df in such functions by the first element only? – user2041143 Mar 03 '13 at 11:50
  • If you think this is rather complicated to understand, you rather start over the whole thinking over a brand new clean paper. – Berci Mar 03 '13 at 11:50
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    The multiplication, addition and subtraction are defined for all real numbers, without any exception (unlike division!), so any expression built by only these is everywhere defined. In particular, $x\cdot x\cdot x+x\cdot x-x-1$ is defined for every $x$. – Berci Mar 03 '13 at 11:53
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    The last comment pretty much explained everything. Thanks a lot for your help! – user2041143 Mar 03 '13 at 11:54