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Math Help - Fuction help!!!

  1. #1
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    Fuction help!!!

    I need some help with function problems. Thanks!!!

    1. find the domain of p(x)=(x^2-2x/x-1) all under a square root sign.
    I was thinking that x could not equal 1,2 the 1 would make the denominator 0 and the 2 would make the numerator 0. So I have no idea because it wasn't the right answer at all.

    2. Given the function f(x)=1/x+1
    find the following expression and simplfy: f(a+h)-f(a)/h

    Given that:
    f(a+h)=1/a+h+1
    f(a)=1/a+1


    3. Find the average rate of change of the function f(t)=(t)^1/2 so t square rooted. between t=a and t=a+t

    I got to here

    ((a)^1/2-(a+h)^1/2)/(a-a+h)

    I know that in the denominator the a's cancel out leaving h, but I have no idea what to do with the top.



    For these problems I need help finding the domain and range. I guess I don't really understand the range either.

    1. f(x)=x^2+1
    what is the Domain and range?

    2. f(x)=x^2+1 given the restriction -1<or equal x <or equal 5
    The domain is [-1,5]
    what is the range?
    is there a simple way to figure this out?
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  2. #2
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    Quote Originally Posted by lucy2284 View Post
    I need some help with function problems. Thanks!!!

    1. find the domain of p(x)=(x^2-2x/x-1) all under a square root sign.
    I was thinking that x could not equal 1,2 the 1 would make the denominator 0 and the 2 would make the numerator 0. So I have no idea because it wasn't the right answer at all.

    2. Given the function f(x)=1/x+1 Mr F says: Is it (1/x) + 1 or 1/(x+1)?
    find the following expression and simplfy: f(a+h)-f(a)/h

    Given that:
    f(a+h)=1/a+h+1
    f(a)=1/a+1


    3. Find the average rate of change of the function f(t)=(t)^1/2 so t square rooted. between t=a and t=a+t Mr F says: Should this be t = a + h?

    I got to here

    ((a)^1/2-(a+h)^1/2)/(a-a+h)

    I know that in the denominator the a's cancel out leaving h, but I have no idea what to do with the top. Mr F says: Do nothing. Becuase there's nothing to do.


    For these problems I need help finding the domain and range. I guess I don't really understand the range either.

    1. f(x)=x^2+1
    what is the Domain and range?

    Mr F says: Draw a graph of this parabola to see what each will be. The range is all possible values of y.

    2. f(x)=x^2+1 given the restriction -1<or equal x <or equal 5
    The domain is [-1,5]
    what is the range?

    Mr F says: Again, draw a graph of this parabola over the given domain to see what each will be. The range is all possible values of y.

    is there a simple way to figure this out?
    1. You require the solution to \frac{x^2-2x}{x-1} \geq 0 where x \neq 1.
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  3. #3
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    re:help for functions

    2. Given the function f(x)=1/x+1 Mr F says: Is it (1/x) + 1 or 1/(x+1)? It is 1/(x+1) sorry about the confusion.
    find the following expression and simplfy: f(a+h)-f(a)/h

    Given that:
    f(a+h)=1/a+h+1
    f(a)=1/a+1

    1. find the domain of p(x)=(x^2-2x/x-1) all under a square root sign.
    I was thinking that x could not equal 1,2 the 1 would make the denominator 0 and the 2 would make the numerator 0. So I have no idea because it wasn't the right answer at all.

    1. You require the solution to where . I was confused by this??? That question was all under a square root.
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  4. #4
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    Quote Originally Posted by lucy2284 View Post
    2. Given the function f(x)=1/x+1 Mr F says: Is it (1/x) + 1 or 1/(x+1)? It is 1/(x+1) sorry about the confusion.
    find the following expression and simplfy: f(a+h)-f(a)/h

    Given that:
    f(a+h)=1/a+h+1
    f(a)=1/a+1

    1. find the domain of p(x)=(x^2-2x/x-1) all under a square root sign.
    I was thinking that x could not equal 1,2 the 1 would make the denominator 0 and the 2 would make the numerator 0. So I have no idea because it wasn't the right answer at all.

    1. You require the solution to where . I was confused by this??? That question was all under a square root.
    The square root can only operate on numbers that are greater than or equal to zero. The square root is operating on \frac{x^2-2x}{x-1}. Therefore ....
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