# domain of f

• Oct 23rd 2006, 07:25 PM
robinpatrick
domain of f
f(x) = (square root -x) / x^3 + 2x^2 - 3x find the domain of f
• Oct 23rd 2006, 07:33 PM
ThePerfectHacker
Quote:

Originally Posted by robinpatrick
f(x) = (square root -x) / x^3 + 2x^2 - 3x find the domain of f

1)The square root needs to be non-negative.
Thus, $\displaystyle \sqrt{-x}$ for $\displaystyle -x\geq 0$ thus, $\displaystyle x\leq 0$.

2)The numerator cannot be zero.
$\displaystyle x^3+2x^2-3x=0$
$\displaystyle x(x^2+2x-3)=0$
$\displaystyle x(x+3)(x-1)=0$
Thus,
$\displaystyle x\not = 0,1,-3$.

Since $\displaystyle x\leq 0$ the only one of those that works is $\displaystyle x=-3$ thus,
$\displaystyle x\leq 0 \mbox{ and } x\not = -3$
Thus,
$\displaystyle (-\infty,-3)\cup (-3,0]$
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[Note for you, for anyone interested in anything I have to say].
These domain seem not be be well-defined, because to definied a function you need to create a domain and co-domain which you did not do here. Thus, how can you find a domain?
• Oct 24th 2006, 06:20 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
1)The square root needs to be non-negative.
Thus, $\displaystyle \sqrt{-x}$ for $\displaystyle -x\geq 0$ thus, $\displaystyle x\leq 0$.

2)The numerator cannot be zero.
$\displaystyle x^3+2x^2-3x=0$
$\displaystyle x(x^2+2x-3)=0$
$\displaystyle x(x+3)(x-1)=0$
Thus,
$\displaystyle x\not = 0,1,-3$.

Since $\displaystyle x\leq 0$ the only one of those that works is $\displaystyle x=-3$ thus,
$\displaystyle x\leq 0 \mbox{ and } x\not = -3$
Thus,
$\displaystyle (-\infty,-3)\cup (-3,0]$
-----
[Note for you, for anyone interested in anything I have to say].
These domain seem not be be well-defined, because to definied a function you need to create a domain and co-domain which you did not do here. Thus, how can you find a domain?

Slight correction. x cannot be 0 else the denominator is zero. So the domain would be:

$\displaystyle (-\infty,-3)\cup (-3,0)$

-Dan