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Math Help - domain, range, and trig

  1. #1
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    domain, range, and trig

    well anyways, I'm working on applying domain and range to trig functions. I had a sheet of about 10 questions and 2 bonus ones. I got one bonus, but not the other and need help with another one on the work sheet.


    Find the domain and range of:

    f(x) = 2sin^-1(3x+4)-1

    g(x) = sqrt[tan^-1(x^2 - 2) - pi/4]


    Thanks for your help!
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  2. #2
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    Quote Originally Posted by Mr_Green View Post
    f(x) = 2sin^-1(3x+4)-1
    Let us begin with the domain.
    The domain of y=asin(x) is [-1,1].

    Thus, for the function,
    f(x)=2asin(3x+4)-1

    We require that,
    -1<=3x+4<=1
    -5<=3x<=-3
    -5/3<=x<=-1

    Now the range.
    To find the range we use my little trick.
    "The range of y=f(x) are all y such that the equation y=f(x) has at least one solution for x in the domain".

    We begin by writing,
    y=2*asin(3x+4)-1
    Thus,
    y+1=2*asin(3x+4)
    Thus,
    (y+1)/2=asin(3x+4)

    In order to have a solution i.e.
    (1/3)sin((y+1)/2)-4=x

    We require that,
    -pi/2<=(y+1)/2<=pi/2
    -pi-1<=y<=pi-1
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  3. #3
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    Quote Originally Posted by Mr_Green View Post
    g(x) = sqrt[tan^-1(x^2 - 2) - pi/4]
    Begin with the domain.

    We notice that atan(x) is always defined for any x.

    However, since there is a square root we have,
    atan(x^2-2)-pi/4>=0
    atan(x^2-2)>=pi/4
    Take tangent of both sides (since the function is increasing).
    x^2-2>=1
    x^2>=3
    x>=sqrt(3) or x<=-sqrt(3)

    Now the range.
    Begin by writing,
    y=sqrt(atan(x^2-2)-pi/4)
    In order to have a solution we require that y>=0 (because otherwise the square root is always positive).
    Thus, if y>=0 then,
    y^2=atan(x^2-2)-pi/4
    y^2+pi/4=atan(x^2-2)
    To take the tangent of both sides, that is,
    tan(y^2+pi/4)=x^2-2

    We require that,
    -pi/2<=y^2+pi/4<=pi/4
    Solving,
    -sqrt(pi)/2<=y<=sqrt(pi)/2
    But we already know that y>=0
    Thus,
    0<=y<=sqrt(pi)/2

    Thus,
    tan(y^2+pi/4)+2=x^2
    To have a solution in the domain we require that,
    The left hand side is positive, which it is, because y>=0.
    Thus, the range is,
    [0,sqrt(pi)/2]
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