Thread: domain of function

1. domain of function

1.) Which real number cannot be in the domain of this function?

f(x) =
_3x-1_________
2x^2 - 5x + 2

2.) Simplify by removing a factor equal to 1:

4a - 6
3 - 2a

3.) Multiply, simplify:

x^3 + 8 X ..x^6 - 4x^5 + 4x^4
x^5 - 4x^3 . . . . .x^2 - 2x + 4

4.) Divide, simplify:

z^2 - 8z + 16 +. . . 3z + 12
z^2 + 8z + 16. . . . . . z^2 - 16

5.)

If f(x) = -2x. . and g(x) = x^2 - 9
. . . . ..x + 3 . . . . . . . . . . . 3 - x .............., write the simplified form of the product function f(g)(X) and identify its domain.

6.)
Identify the missing numerator in the statement below that would make the equation TRUE:

x - 20............-....?????????.........=.... 3
x^2 - 8x + 12.........x^2 - 8x + 12........x-2

7.)
Add, simplify the results:

4s + ...........s
s^2 - 16....s + 4

Sorry that its hard to read, but I'm fairly new to this and I was having trouble with it bunching together when I would use my 'space' key. Help would be very much appreciated!

Thanks in advance.

2. Originally Posted by zagsfan20
1.) Which real number cannot be in the domain of this function?

f(x) =
_3x-1_________
2x^2 - 5x + 2
(3x - 1)/[(x - 2)*(2x - 1)]

You do not want a 0 in the denominator;

Therefore,

(x - 2)*(2x - 1) = 0

x = 2 V x = 1/2,

Thus, both of the above x values cannot be in the domain of the function.

3. Originally Posted by zagsfan20
2.) Simplify by removing a factor equal to 1:

4a - 6
3 - 2a
I'll do one more since I have to be up early tomorrow and its 3 AM .

Any way, I'll help you 'remove a factor equal to 1,' where you should be able to solve from there.

(4a - 6)/(3 - 2a)

Is equivalent to:

[2*(2a - 3)]/[2*(-a + 3/2)]

2/2 is just 1;

(2a - 3)/(-a + 3/2); you should be able to take it from here.

And I'm sure others you'll get prompt replies on your other questions soon.

4. 3. $\frac{x^3+8}{x^5 - 4x^3} \times \frac{x^6 - 4x^5 + 4x^4}{x^2 - 2x + 4}$

$= \frac{(x + 2)(x^2 - 2x + 4)}{x^3(x^2 - 4)} \times \frac{x^4(x^2 - 4x + 4)}{x^2 - 2x + 4}$ <-- Cancel the $x^2 - 2x + 4$ and an $x^3$

$= \frac{x + 2}{x^2 - 4} \times \frac{x(x^2 - 4x + 4)}{1}$

$= \frac{x + 2}{(x + 2)(x - 2)} \times \frac{x(x - 2)^2}{1}$ <-- Cancel the $x + 2$ and one $x - 2$

$= \frac{1}{1} \times \frac{x(x - 2)}{1}$

$= x(x - 2)$

We should state that $x \neq 0$ and $x \neq \pm 2$ to indicate the terms we cancelled. (Typically such problems only involve real numbers and since there is no real number x such that $x^2 - 2x +4 = 0$ we may ignore that we cancelled this in stating the domain of the expression.)

-Dan

5. Originally Posted by zagsfan20
4.) Divide, simplify:

z^2 - 8z + 16 +. . . 3z + 12
z^2 + 8z + 16. . . . . . z^2 - 16
Given the instructions I'm not sure I understand this. Are we doing:
$\frac{z^2 - 8z + 16}{z^2 + 8z + 16} + \frac{3x+12}{z^2 - 16}$

-Dan

6. Originally Posted by zagsfan20
5.)

If f(x) = -2x. . and g(x) = x^2 - 9
. . . . ..x + 3 . . . . . . . . . . . 3 - x .............., write the simplified form of the product function f(g)(X) and identify its domain.
$f(x) = \frac{-2x}{x + 3}$ and $g(x) = \frac{x^2 - 9}{3 - x}$

$f(g(x)) = f \left ( \frac{x^2 - 9}{3 - x} \right )$

Now, to make things a bit easier I'm going to write the argument here as:
$\frac{x^2 - 9}{3 - x} = \frac{(x + 3)(x - 3)}{-(x - 3)}$ <-- Cancel the $x - 3$

$= -(x + 3)$.

BUT we have to remind ourselves that $x \neq 3$.

So
$f(g(x)) = f \left ( \frac{x^2 - 9}{3 - x} \right ) = f(-(x + 3))$, $x \neq 3$

$= \frac{-2 [-(x + 3)]}{[-(x+3)] + 3}$, $x \neq 3$

$= \frac{2x + 6}{-x - 3 + 3}$, $x \neq 3$

$= \frac{2x + 6}{-x}$, $x \neq 3$

$= - \frac{2x + 6}{x}$, $x \neq 3$

Now, note that $x \neq 0$ as well. Thus the domain is $( -\infty, 0) \cup (0 , 3) \cup (3, \infty)$

-Dan

7. Originally Posted by topsquark
Given the instructions I'm not sure I understand this. Are we doing:
$\frac{z^2 - 8z + 16}{z^2 + 8z + 16} + \frac{3x+12}{z^2 - 16}$

-Dan
Yeah, that what I was trying to post. I'm fairly new to this how do you get the capability to write out problems like that?

8. Originally Posted by zagsfan20
4.) Divide, simplify:

z^2 - 8z + 16 +. . . 3z + 12
z^2 + 8z + 16. . . . . . z^2 - 16
z^2 - 8z + 16 = (z - 4)^2

z^2 + 8z + 16 = (z + 4)^2

3z + 12 = 3(z + 4)

z^2 - 16 = (z - 4)(z + 4)

Thus, we have:

[(z - 4)^2/(z + 4)^2] + [(3(z + 4))/((z - 4)(z + 4))]

We'll work with the part in italics above;

It's clear that we can cancel a z + 4, and we have:

[(z - 4)^2/(z + 4)^2] + 3/(z - 4)

You could expand the [(z - 4)^2/(z + 4)^2], or keep it the way it is, whichever you prefer.

9. Originally Posted by zagsfan20
7.)
Add, simplify the results:

4s + ...........s
s^2 - 16....s + 4
I'll do #7, since I assume Dan is doing #6.

(4s)/(s^2 - 16) + s/(s + 4)

Factor s^2 - 16;

(s - 4)*(s + 4)

(4s)/[(s - 4)*(s + 4)] + s/(s + 4)

Find a common denominator; multiply the two denominators and determine the numerator;

((x + 4)*(4s))/[(x - 4)*(x + 4)*(x + 4)]

(4s^2 + 16s)/[(x - 4)*(x + 4)^2] + [((x - 4)*(x + 4))*x]/[(x - 4)*(x + 4)^2]

Now, we have a common denominator. Yay!

Now add the numerator and divide it by the same denominator;

[x^2*(x + 4)]/[(x - 4)*(x + 4)^2]

Cancel a (x + 4);

Thus, the final answer: x^2/[(x - 4)*(x + 4)]

10. Originally Posted by zagsfan20
Yeah, that what I was trying to post. I'm fairly new to this how do you get the capability to write out problems like that?
See the LaTeX tutorials at the beginning of this thread.

Originally Posted by AfterShock
I'll do #7, since I assume Dan is doing #6.
Actually I'm not. I've got too many posts to go through. New girlfriends are distracting and time consuming. (But worth the effort! )

-Dan

11. Originally Posted by zagsfan20
6.)
Identify the missing numerator in the statement below that would make the equation TRUE:

x - 20............-....?????????.........=.... 3
x^2 - 8x + 12.........x^2 - 8x + 12........x-2
(x - 20)/(x^2 - 8x + 12) - Y/(x^2 - 8x + 12) = 3/(x - 2)

They made this easy by giving you the common denom;

Factor x^2 - 8x + 12;

(x - 6)*(x - 2)

Whatever is in the numerator, Y, we want the result to be 3*(x - 6) so we can cancel an (x - 6) and be left with 3/(x - 2)

So, (x - 20) - {expression here} = 3*(x - 6)

(x - 20) - {expression here} = 3*x - 18

Thus, we have

(x - 20) - (-2x - 2);

Plug the bold part in for Y and check that it works;

(x - 20)/(x^2 - 8x + 12) - (-2x - 2)/(x^2 - 8x + 12) = 3/(x - 2)

[(x - 20) - (-2x - 2)]/[(x^2 - 8x + 12)]

(3x - 18)/[(x - 6)*(x - 2)]

3(x - 6)/[(x - 6)*(x - 2)]

Cancel the (x - 6);

Thus, 3/(x - 2), which is what you were looking for.