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March 26, 2017

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The point (5,-2) lies on the graph of y=f(x), and assume f^-1(x) exists. Find the corresponding point on eAch graph.

A). Y=f^-1(x)

I get (-5,2)


B). Y=-f^-1(-x)+4
I get (5,2)



C). Y=f^-1(x+2)+3

I get (-7,5)


Did I get these right?

  • precalc!!!!! - ,

    say y = x - 7
    (5,-2) lies on that graph
    for f^-1
    x = y - 7
    y = x + 7
    if x = 5 then y = 12
    so I get
    (5,12)

  • dumb mistake - ,

    say y = x - 7
    (5,-2) lies on that graph
    for f^-1
    x = y + 7
    y = x - 7
    if x = 5 then y = -2
    so I get
    (5,-2)

  • precalc!!!!! - ,

    B. My f^-1(x) = x-7
    Y=-f^-1(-x)+4
    Y = (-5-7) + 4
    Y = -12+ 4 = -8
    (5,-8)

    C.Y=f^-1(x+2)+3
    x+2 = 7
    f^-1(7) = 7-7 = 0
    Y = 0 + 3 = 3

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