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Math Help - inverse functions

  1. #1
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    inverse functions

    f(x)=-4x-2 & g(x)=5x-6. Find (f*g) and state domain.


    f(x)-x^2-1 what is the equation for f^-1(x)?


    f(x)=3x+2, find f(f^-1(14)).


    f(x)=4x+7 and g(x)=3x-5 find (f*g)(-4)


    f(x)=sqrt x+3 what is equation for f^-1(x)?


    Graph y=sqrt x-2 +5 which point lies on graph?
    A(7,6) B(0,5) C(-2,5) D(3,6)

    The graph y=sqrt x is shifted 6 points down and 1 unit right. Which equation represents the new graph.
    y=sqrt x+1 -6
    y=sqrt x-6 +1
    y= sqrt x-1 -6
    y= sqrt x+6-1
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  2. #2
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    Quote Originally Posted by pendulum View Post
    f(x)=-4x-2 & g(x)=5x-6. Find (f*g) and state domain.
    f*g = (4x - 2)(5x - 6)

    now multiply and sinplify.

    Domain = x \in \mathbb{R}
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  3. #3
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    Quote Originally Posted by pendulum View Post
    f(x)-x^2-1 what is the equation for f^-1(x)?
    f(x) = - x^2  - 1

    y = - x^2  - 1

    interchange x and y and solve for y

    x = - y^2  - 1

     <br />
y^2 = - x - 1<br />

     <br />
y = \pm \sqrt{ - x - 1}<br />

     <br />
f^{-1}(x)= \pm \sqrt{ - x - 1}<br />

    in the same way do the following parts and show your work

    f(x)=3x+2, find f(f^-1(14)). , find f^{-1}(x) in the same way, then put x = 14 and calculate


    f(x)=4x+7 and g(x)=3x-5 find (f*g)(-4) , calculate f*g and simplify, then put x = -4 and calculate


    f(x)=sqrt x+3 what is equation for f^-1(x)? find f^{-1}(x) in the same way
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  4. #4
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    Quote Originally Posted by pendulum View Post
    Graph y=sqrt x-2 +5 which point lies on graph?
    A(7,6) B(0,5) C(-2,5) D(3,6)
    Put the point in eqn and see if it satisfies it or not.

    y= \sqrt {x-2} +5

    for first point, (7, 6), y= \sqrt {7-2} +5

    y= \sqrt {5} +5 \ne 6

    for fourth point, (3, 6), y= \sqrt {3-2} +5

    y= \sqrt {1} +5 = 6

    so point (3, 6) lies on graph.

    Please see attached graph
    Attached Thumbnails Attached Thumbnails inverse functions-g3.jpg  
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  5. #5
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    Quote Originally Posted by pendulum View Post
    The graph y=sqrt x is shifted 6 points down and 1 unit right. Which equation represents the new graph.
    y=sqrt x+1 -6
    y=sqrt x-6 +1
    y= sqrt x-1 -6
    y= sqrt x+6-1
    If the graph of y = \sqrt{x} is shifted 6 units down, then eqn will be

    y = \sqrt{x} - 6

    If this is moved 1 unit right, then eqn will be

    y = \sqrt{x-1}-6

    Did you get it ???
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