sets ,domain, range and functions.

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November 3rd 2006, 07:38 AM
m777

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sets ,domain, range and functions.

please try to solve these questions.
November 3rd 2006, 07:52 AM
topsquark

Quote:

Originally Posted by m777

please try to solve these questions.

2. Let $f: \mathbb{Z} \to \mathbb{Z}: n \mapsto n^2$

This function is not one to one. If f were one to one then there would exist only one $n^2$ for every n. But $n_1 = -x$ and $n_2 = x$ where x is a positive integer both share the same value in the range of f: $f(n_1) = f(n_x) = x^2$ . So f is not one to one.

-Dan
November 3rd 2006, 08:00 AM
ThePerfectHacker

1)Yes the ordering relation on the reals is an equivalence relations. Thus it is: reflesive, symettric, transitive.

3)Domain is all reals except -5.

The range all reals $y$ such as,
$y=\frac{x-1}{x-5}$
Has solution in the domain.
Thus,
$y=\frac{(x-5)+4}{x-5}$
$y=1+\frac{4}{x-5}$
$y-1=\frac{4}{x-5}$
If, $y-1\not = 0$
Then,
$\frac{5}{y+1}+5=x$
Has a solution in the domain.
If, $y=1$
Then,
$0=\frac{4}{x-5}$
Which has no solutions.
Thus the range is,
$(-\infty,1)\cup (1,\infty)$
November 3rd 2006, 08:04 AM
topsquark

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Quote:

Originally Posted by m777

please try to solve these questions.

Find the domain and range of $f(x) = \frac{x-1}{x-5}$ .

The domain is easy. Find all values of x such that:
1) The denominator is 0.
2) The function under a square root is negative.
These are the usual things to worry about. These values of x are not in the domain of f(x).

In this case all we need to worry about is the denominator. So where is x - 5 = 0? x = 5. So this is the only point not in the domain of f(x). Thus the domain is: $(-\infty, 5) \cup (5, \infty)$ .

The range is a bit trickier, but simple enough to see in this case. ThePerfectHacker will likely poke his nose in with his method, but for now look at the graph of the function which I have attached. It clearly shows that the function spans from $(-\infty, infty)$ . The only question is whether the function is ever equal to 1, where I have sketched a dotted line for the horizontal asymptote. So can we solve:
$1 = \frac{x-1}{x-5}$

$x - 5 = x - 1$

$-5 = -1$
Which obviously is not true. So f(x) cannot be 1. Thus the range is $(-\infty,1) \cup (1, \infty)$ .

(Edit: I see TPH beat my post. However he forgot a negative sign in his equation. The range above is correct.)

-Dan
November 3rd 2006, 08:04 AM
ThePerfectHacker

4)
$f=\sqrt{x+1}$
$g=1/x^2$
Thus,
$f\circ g=\sqrt{\frac{1}{x^2}+1}=\sqrt{\frac{1+x^2}{x^2}}= \frac{\sqrt{1+x^2}}{|x|}$

$g\circ f=\frac{(\sqrt{x+1})^2}=\frac{1}{x+1}$
November 3rd 2006, 10:16 AM
Soroban

Hello, m777!

Here's my approach to #4 . . .

Quote:

Given: . $f(x) = \sqrt{x+1},\;g(x) = \frac{1}{x^2}$

Find: . $f\circ g$ and $g\circ f$

$f\circ g \;= \;f(\underbrace{g(x)}_\downarrow)$
. . . . $= \;f\left(\frac{1}{x^2}\right)$

. . . . $= \;\sqrt{\frac{1}{x^2} + 1} \;=\;\sqrt{\frac{1+x^2}{x^2}} \;= \;\frac{\sqrt{1+x^2}}{|x|}$

$g\circ f \;= \;g(\underbrace{f(x)}_\downarrow)$
. . . . $= \;g\left(\sqrt{x+1}\right)$

. . . . $= \;\frac{1}{(\sqrt{x+1})^2} \;= \;\frac{1}{x+1}$