Results 1 to 6 of 6

Math Help - y = f(x) = (x-5)

  1. #1
    Member
    Joined
    Sep 2006
    Posts
    99

    y = f(x) = (x-5)

    y = f(x) = (x-5)

    Write an equation which represents x = f(y) and solve for y.

    How to solve this? Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by shenton View Post
    y = f(x) = (x-5)

    Write an equation which represents x = f(y) and solve for y.

    How to solve this? Thanks.
    Hello,

    if I understand your question correctly you should do:

    x=(y-5)^2 \Longleftrightarrow \sqrt{x}=y-5\ \vee \ -\sqrt{x}=y-5 \ \Longleftrightarrow 5+\sqrt{x}=y\ \vee \ 5-\sqrt{x}=y

    EB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2006
    Posts
    99
    Thanks, I don't understand the question myself. I copied direct from the paper.

    The topic heading is "Comparing the Graphs of y=f(x) and y=f-1(x) or x=f(y)"

    y=f-1(x) is the inverse of y=f(x).

    I don't know if x=f(y) is the same as y=f-1(x) or if it is a different thing.

    Your answer makes sense. We know y = (x-5) produce a "U" graph that was moved 5 units to the right.

    This answer, f-1(x) = √(x) + 5, should produced a "U" shape rotated at 180 degrees but since the graph of radical functions has x-values > 0, I do not see the "U' shape.

    Just some comments, sorry, I don't know what I'm taliking since I am not an expert in inverse functions of a quadratic expression.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by shenton View Post
    Thanks, I don't understand the question myself. I copied direct from the paper.

    The topic heading is "Comparing the Graphs of y=f(x) and y=f-1(x) or x=f(y)"

    y=f-1(x) is the inverse of y=f(x).

    I don't know if x=f(y) is the same as y=f-1(x) or if it is a different thing.

    Your answer makes sense. We know y = (x-5) produce a "U" graph that was moved 5 units to the right.

    This answer, f-1(x) = √(x) + 5, should produced a "U" shape rotated at 180 degrees but since the graph of radical functions has x-values > 0, I do not see the "U' shape.

    ...
    Hello,

    you wrote ...should produced a "U" shape rotated at 180 degrees.... Not quite, because then you get a parabola opened downward but not the graph of the inverse function. I attached a diagram with the original function and both inverse function of parts of f.
    Similar colours mean that the corresponding graphs are reflected at the line y = x (and that's by the way the method you can "construct" point for point the inversed graph) To control your drawings (or sketches): If the graph of a function intersects with the line y = x then the graph of the inverse function must do so too.

    EB
    Attached Thumbnails Attached Thumbnails y = f(x) = (x-5)²-graph_inversfkt.gif  
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2006
    Posts
    99
    You are right about the graph of the inverted function.

    If the graph of a function intersects with the line y = x then the graph of the inverse function must do so too - Something new for me, that is good to know.

    We worked out the inverse of y = (x-5) is y-1 = √(x) + 5. I know the graph is correct since the vertex of the function is (0,5) and the vertex of the inverse function is (5,0).

    What is strange is that √(x) + 5 will not produce the "U" shape inverted graph. Algebratically, it looks right. Graphically, it looks right. I am stumped now.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by shenton View Post
    ...

    What is strange is that √(x) + 5 will not produce the "U" shape inverted graph. Algebratically, it looks right. Graphically, it looks right. I am stumped now.
    Hello,

    it's me again.

    The U you are looking for opens to the right {} \subset {} and is composed of 2 parts:
    y = √(x) + 5 is the upper part because you add a value to 5

    y = -√(x) + 5 is the lower part because you subtract a value from 5.

    You can easily detect the u-shaped curve in my diagram.

    EB
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum