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Math Help - Some problems

  1. #1
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    Some problems

    Find the domain and range of the function

    y = 5 - square root of(9 - x^2)

    i got the domain
    But for the range, my work took a lot of space and came out to something weird. Can anyone work through this and show me how to get the range?

    10. Let f(x) = square root of (3 - x).
    Find an expression for f inverse x
    I got that as y = 3 - x^2

    But it says(Be sure to state any necessary domain restrictions???)

    11. f(x) = cube root of (x + 2) and g(x) = xcubed - 2

    True or false

    a. g(x) = f inverse of x
    i got as True
    b. f of g of x = 1
    i got as True
    c. The function is one to one
    Dont know how to prove this other than graphing and using vert line test
    Is there easier way?


    20. Express y as a function of x, given that the constant C is a positive number : e ^(y + C) = x^2 + 4

    No idea ....lol never did this in last year class

    23. Solve each equation without using a calc

    a. 2lnx - ln3 = 2
    b. ln(x - 2) + ln(x + 5) = 2ln3

    For these two, i am not remembering something because these should be easy...im not seeing something here...
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  2. #2
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    Quote Originally Posted by ruthvik View Post
    Find the domain and range of the function

    y = 5 - square root of(9 - x^2)

    i got the domain
    But for the range, my work took a lot of space and came out to something weird. Can anyone work through this and show me how to get the range?
    Do you know what shape this function has? I know that would help me find the range.

    Spoiler:
    A circle has the form y^2+x^2 = R^2 where R is the radius



    Quote Originally Posted by ruthvik View Post

    10. Let f(x) = square root of (3 - x).
    Find an expression for f inverse x
    I got that as y = 3 - x^2

    But it says(Be sure to state any necessary domain restrictions???)

    The inverse is correct, you need to restrict the domain of your function so it is 1 to 1.



    Quote Originally Posted by ruthvik View Post
    11. f(x) = cube root of (x + 2) and g(x) = xcubed - 2

    True or false

    a. g(x) = f inverse of x
    i got as True
    b. f of g of x = 1
    i got as True
    c. The function is one to one
    Dont know how to prove this other than graphing and using vert line test
    Is there easier way?
    I agree with a)

    for b) I think f[g(x)]=x

    c) Both functions are 1 to 1. Vertical line test is always a winner! Easier way is to learn the shape of these functions.


    Quote Originally Posted by ruthvik View Post

    20. Express y as a function of x, given that the constant C is a positive number : e ^(y + C) = x^2 + 4

    No idea ....lol never did this in last year class
    e ^{y + C} = x^2 + 4

    y + C = \ln(x^2 + 4)

    y = \ln(x^2 + 4)-C


    Quote Originally Posted by ruthvik View Post
    23. Solve each equation without using a calc

    a. 2lnx - ln3 = 2
    b. ln(x - 2) + ln(x + 5) = 2ln3

    For these two, i am not remembering something because these should be easy...im not seeing something here...

    2\ln(x) - \ln(3) = 2

    \ln(x^2) - \ln(3) = 2

    \ln(\dfrac{x^2}{3}) = 2

    \ln(\dfrac{x^2}{3}) = 2\ln(e)

    \ln(\dfrac{x^2}{3}) = \ln(e^2)

    \dfrac{x^2}{3} =e^2

    Can you finish it from here?

    \ln(x - 2) + \ln(x + 5) = 2\ln(3)

    \ln((x - 2) (x + 5)) = 2\ln(3)

    \ln((x - 2) (x + 5)) = \ln(3^2)

    (x - 2) (x + 5) = 3^2

    x^2+3x-10 = 9

    x^2+3x-19 = 0

    Can you finish it from here?
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  3. #3
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    So the first one is half a circle?
    looks like a circle almost when i graph it but is restricted


    For number 10, what would the restrictions be? how did u arrive at yours? graph it?


    For the last two

    i got up to those parts

    wasnt sure of b because it didnt factor but ill just find in decimals

    and no idea how to do 23. a. even with that

    it says no calc


    I agree with everything else u did though, thanks a lot
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  4. #4
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    Quote Originally Posted by ruthvik View Post
    So the first one is half a circle?
    looks like a circle almost when i graph it but is restricted
    The negative in front of the square restricts it to a semicircle


    Quote Originally Posted by ruthvik View Post

    For number 10, what would the restrictions be? how did u arrive at yours? graph it?
    I found the inverse by swapping x and y, then solving for y.


    Quote Originally Posted by ruthvik View Post


    For the last two

    i got up to those parts

    wasnt sure of b because it didnt factor but ill just find in decimals

    and no idea how to do 23. a. even with that

    it says no calc
    \dfrac{x^2}{3} =e^2

    x^2 =3e^2

    x =\sqrt{3e^2}

    Done...

    x^2+3x-19 = 0

    For ax^2+bx+c=0

     x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}
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  5. #5
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    i got range as y >= 2 and less than or equal to 4.5?

    Also, for number 10, what are the restrictions?
    Last edited by ruthvik; September 1st 2009 at 09:57 AM.
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