Domain of Function;Graphing;Zeros algebraically
I have four problems I need help with.
The first two are
Find the domain of the function:
√25x^{2 }Edit: After attempting this problem,I've gotten (5,5) ..is that right?
and
Find the domain of the function:
x/x^{2}x6
Edit: For this one I've gotten (∞,2)U(2,3)U(3,∞).Once again,I'd appreciated some double checking from someone.
The next is
Find the zeros of the function algebraically:
f(x)=x^{3}x^{2}25x+25
Edit: And for this one,I've gotten 5,5,1.Correct?
The last is
Graph the function:
x^{2}2, x<2
f(x)= 5 , 2< x <0
8x5, x >0
^This one specifically I really need help understanding. ):
Re: Domain of Function;Graphing;Zeros algebraically
For your first check, the √25x2 is not limited in any domainas it is a parabola opening downward. If graphing is not available simply look at the equation... is there any value of x that would not allow it to be a real number? because x can only be squared, there are no limits.
Again, with your second problem there are no limits on what x can be (which is what a domain is) except that it cannot be zero because if it was zero, then the fraction would become undefined. That would get you a domain of all reals excluding zero because zero squared would equal zero.
Your third is correct :)
This last one is called a piecewise function, where each segment if the function has the properties of a different toolkit. so, to help you understand, i will write out how that equation would sound. "y will equal X squared minus 2 when X is less than negative 2". that is the first part. So wherever x is less than 2, you would graph x^22, and then stop at 2 with an open circle because it cannot actually equal 2. The next would read like "F of X equals 5 (Y=5) when x is greater than or equal to 2 and less than of equal to 0" same deal. Finally, " will equal 8x minus 5 when x is greater than zero (not including zero)" to symbolize an end that includes the final value is a closed circle.
Hope this has helped
Re: Domain of Function;Graphing;Zeros algebraically
My understanding was that the restriction for the first problem is anything that results in a negative root,because a negative root isn't possible (without getting into imaginary numbers).And for the second,my reasoning was that when 2 is plugged into the denominator you get 4+26,which would be zero and when you plug in 3 you get 936,which again is a zero;while anything below 2,above 3,and inbetween excluding those numbers would result in either a negative denominator or a positive one,hence my interval notation.
I believe your last explanation does help me understand better (especially where the closed/open circles are concerned) and I'm trying very hard to understand,so thank you for replying back!
Re: Domain of Function;Graphing;Zeros algebraically
Re: Domain of Function;Graphing;Zeros algebraically
Quote:
Originally Posted by
Plato I think that FxAperture read the question as
and not as
.
That is because you failed to use any grouping symbols at all in the entire post.
For example x/x^2x6 should be read as
, but you meant
. So group things.
Oh,okay.Sorry,I'm not very literate when it comes to how to type out equations.Thank you.
Re: Domain of Function;Graphing;Zeros algebraically
Hello, Ai Ekio!
FxAperture did an excellent job explaining the piecewise function.
I'll babystep through its graphing . . .
The first function is a parabola: ,
Code:

*  *



*  *


*  *

*  *
*+
*  *
*

But we want only the portion less than 2.
Code:

*  .



*  .


o  .
: 
:.  .
+.+.
2 .  .
.


o

The second function is: . , a horizontal line.
Code:


* * * * * * * * *




+


But we want the portion between 2 and 0, inclusive.
Code:


********
: 
: 
: 
: 
++
2 

The third function is a line.
Code:

 *

 *

 *

 *

 *

+*

*

*

*

But we want the portion greater than zero.
Code:

 *

 *

 *

 *

 *

+*

*

o



Therefore, the graph looks like this:
Code:

*  *

 *

*  *
********
 *
o 
:  *
: 
++*
2 
*

o

Re: Domain of Function;Graphing;Zeros algebraically
Quote:
Originally Posted by
Soroban Hello, Ai Ekio!
Thank you!The visuals really helped. (: