Thread: Domain of Function;Graphing;Zeros algebraically

I have four problems I need help with.

The first two are-
Find the domain of the function:
√25-x²
Edit: After attempting this problem,I've gotten (-5,5) ..is that right?

and
Find the domain of the function:
x/x²-x-6
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)=x³-x²-25x+25
Edit: And for this one,I've gotten 5,-5,1.Correct?

The last is-
Graph the function:

x²-2, x<-2

f(x)= 5 , -2< x <0

8x-5, x >0

^This one specifically I really need help understanding. ):

For your first check, the √25-x2 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^2-2, 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

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+2-6,which would be zero and when you plug in 3 you get 9-3-6,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!

Originally Posted by Ai Ekio

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+2-6,which would be zero and when you plug in 3 you get 9-3-6,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!

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^2-x-6 should be read as , but you meant . So group things.

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^2-x-6 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.

Hello, Ai Ekio!

FxAperture did an excellent job explaining the piecewise function.
I'll baby-step 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
              |

Originally Posted by Soroban

Hello, Ai Ekio!

Thank you!The visuals really helped. (: