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If f(x) = x + cos x, find the inverse of f(1)?

Discussion in 'Mathematics and Physics' started by ram1cal, Mar 12, 2009.

  1. ram1cal

    ram1cal New Member

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    if it makes things clearer, it didn't say inverse, it showed f^-1(1), which pretty much means f prime

    please show steps? this is the only weird one on my homework
     
  2. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Your function is:

    f(x)=x+cos(x)

    and it looks like you need to find f^-1(1) which would mean if we
    had the inverse function x=f^-1(y) we would calculate x knowing y.

    The function can be written:

    y=x+cos(x)

    and since we know what y is we plug that in:

    1=x+cos(x)

    Now we dont know what the inverse function is and it doesnt look
    easy to calculate or it doesnt even exist, so we look at the equation
    above and try to solve it. We need to find an x that when used in
    "x+cos(x)" it comes out to an answer of 1.
    The only solution i can see is x=0, because this is true:
    1=0+cos(0)

    so that means that f^-1(1)=0 because when y equals one x equals zero.
     
    • Like Like x 1
  3. j.friend

    j.friend New Member

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    I'm sorry but that doesn't seem entirely correct. Whilst the inverse may be hard to find, the value of the inverse function cannot be found be subbing 1 into the normal function. Whilst the substituion method may not be able to be utilised, the graphical analysis method may.

    I'm sure you know that the inverse of a function is the function reflected over the line y = x. Either you can make a rough sketch by hand or use a graphical calcultor (common in most schools nowadays for senior mathematics courses).

    By sketching the graph of f(x) = x + cosx, it can be seen that it will have an inverese function due to the graph being 1:1 (meaning that each x point has only one corresponding y point and vice versa).

    Another aspect of inverse functions is that essentially the x and y values swap for the inverse. This means that for the numbers subbed in (if y = 1, x = 0) on the inverse graph there will be a point (0,1) or at x = 0, y = 1.

    Taking this into account by finding the value of the original function when x = 1, the value for the inverse function when y = 1 will be found.

    so y = x + cosx
    if x = 1, y = 1 + cos 1 (your calculator will need to be in radians mode to calculate this properly)

    from this you get y ≈ 1.54

    this means that the value of f^-1(1) ≈ 1.54
     
  4. dave

    Dave New Member

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  5. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi

    Ummm, and how exactly are you utilizing the inverse function there?
    I dont see any inverse function in your solution...did i miss something?

    The original function is:

    y=x+cos(x)

    or, stated another way:

    y=f(x)

    The problem was to find the value of the *inverse* function when
    y is made equal to 1, not to simply substitute 1 in for x. That is,
    if we knew what f' was we could then (and only then) plug 1 into
    it and get the answer.
     
  6. j.friend

    j.friend New Member

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    ah yes, :eek:

    I misread/misinterpretted what was going on. Rather silly of me.

    I apologise
     
  7. Ariek

    Ariek New Member

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    Mr. Al, thanks so much! I had same hw question, but how would I then find deriv of g(1)?
     
  8. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    It looks like part of this thread is missing, the very beginning. It's also around 8 years old :)
     
  9. Ratchit

    Ratchit Well-Known Member

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    Sometimes a plot helps. As you can see, there is only one solution to the inverse of the function.

    MrAl.JPG

    Ratch
     
  10. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello,

    Yes a plot helps, but in this case you did not say how it helps. For example, are you saying that there are an infinite number fo solutions or just one? In other words, you need to state what your final solution to this problem would be.
    We both know what it is, but someone else might not know what you are implying with the graph.
    BTW it's a little clearer to graph y=f(x)-y1 rather than y=f(x).

    Back to main ideas...

    The derivative of the inverse function can be found using the property that the inverse function slope is 1/m where m is the slope of the original function. Watch the sign, and probably you want the derivative at the solution point.
     
    Last edited: Mar 29, 2017
  11. Ratchit

    Ratchit Well-Known Member

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    Where does it say that the derivative is involved anywhere? I thought the remark by the OP about f ' was a misinterpretation. If you think otherwise, would you please restate the problem more understandably? Until I see otherwise, I am assuming the problem is asking for a value when the the inverse of f(x), call it g(x), is g(1). As you observed, it is easily seen by inspection that g(1) = 0. Furthermore, the plot confirms that zero is the only solution to g(1), and shows there is a unique y-value for each x-value and vice-versa. I don't know what more can be said about this.

    Ratch
     
  12. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi Ratch,

    In post #6 it was asked how to find the "deriv" of g.

    What you say now is true, but before that you did not explain what the graph was doing for us. People who dont do this that much dont know what they are looking at. For example, in your graph do we look for the y axis crossing or the x axis crossing or consider all the points on that graph to mean something. Until you explain it someone might not know. Now that you've explained it it will be clear to almost anyone. You have to keep in mind you might be dealing with students who are new to this. I think it is very good to show the graph though, with a short added explanation.

    There are some other interesting ways to solve this too, but not sure if i want to get into that right now.
     
  13. Ratchit

    Ratchit Well-Known Member

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    OK, the plot below is the function f(x).
    MrAl1.JPG

    The next plot in the inverse of f(x) which I will call g(x).

    MrAl2.JPG

    The last plot is the derivative of g'(x)

    MrAl3.JPG

    Regards,

    Ratch
     
  14. kalawi_tan

    kalawi_tan New Member

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    Woah! This is awesome! I missed my college days with Algebra, Physics, Advaced Mathematics, Thermodynamics, and such.. :D
     

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