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Homework Help: Need to show function {f(x)= x + cos x} has an inverse w/out the use of a graph

  1. Apr 8, 2005 #1
    I need to show that the function {f(x)= x + cos x} has an inverse without the use of a graph. I understand that the simplest way to do this is by the using the horizontal line test, but that is not an option in this case. The professor has asked us to think intuitively about this problem, and I am just at a loss. Any help would be most appreciated, as I am trying not to fail this class. Thanks!
  2. jcsd
  3. Apr 8, 2005 #2


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    For a continuous function, what has to be true about the slope for it to pass the horizontal line test? Can there be any maxima or minima?
  4. Apr 8, 2005 #3
    Thanks for the reply. That was really fast.

    I think that the slope must be constant (positive or negative) in order for a function to pass the horizontal line test. I mean, it can be positive AND negative. I also believe that there cannot be any maxima or minima.

    I don't want to seem unintelligent, but I am not so great at calculus to begin with, and my teacher this semester if fresh out of grad school. He is a really nice person, but he doesn't always seem to understand that we don't know everything he does. So, on some topics his instruction is sketchy at best. But, there's nothing I can do about that, so I am making the best of this situation.
  5. Apr 8, 2005 #4


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    Right, it must either always be positive or always be negative. That doesn't mean it's constant, though, which would be a line. So just make sure the first derivative of your function always has the same sign.
  6. Apr 8, 2005 #5
    I am not sure what you mean by this. The 1st derivative of my function is (I think) f'(x)=sin x .

    I know f(x)= x^2 does not have an inverse. The first derivative here is f'(x)=2x

    I know f(x)=e^x does have an inverse. The first derivative here is f'(x)=lnx.

    I just don't know how all of this is related. Is there a rule (or fundamental concept) I am missing?
  7. Apr 8, 2005 #6


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    Your derivative is wrong. Don't forget the derivative of x is 1. And the derivative of e^x is e^x. x^2 doesn't have an inverse because its derivative is negative for x<0 and positive for x>0. e^x does because it is always positive.

    It's pretty intuitive if you think about what happens when a derivative changes sign. It means there is a maximum or minimum, and the function will fail the horizontal line test.
    Last edited: Apr 8, 2005
  8. Apr 8, 2005 #7
    the first derivative tells you when the graph is increasing or decreasing, and for a function to have an inverse it has to be always increasing or decreasing if you think about it it will make sense. hence, when the derivative is positive the graph is increasing and when the derivative is negative the graph is decreasing. So using logic for a function to have an inverse its derivative has to be either always positive or negative. the derivative of your function is simple enough to compute
    f'(x)= 1- sin(x) Can you figure out why this is either always positive or negative?
  9. Apr 8, 2005 #8

    "the derivative of your function is simple enough to compute
    f'(x)= 1- sin(x) Can you figure out why this is either always positive or negative?"

    I can't, can you?
  10. Apr 8, 2005 #9
    yes if you wish to know the answer send me a message, I thought that it would be too easy for him if i just give him the answer.
  11. Apr 8, 2005 #10

    Weren't there some bounds given for the domain of f(x). If not, you're going to have a hard time showing the f(x) has an inverse!
  12. Apr 8, 2005 #11
    No, no bounds were given for the domain of x.

    Also, I am not 100% sure why f'(x)= 1- sin(x) is always positive or negative. Perhaps I have not had enough experience with derivatives for this to be evident immediately.
  13. Apr 8, 2005 #12


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    -sin x is always greater than or equal to -1. now add 1 to it.
  14. Apr 8, 2005 #13
    sin x has a range of -1 to 1, that means at its lowest the value sin will be -1 and its highest 1. Input those values in your derivative eq. 1-(-1)=2 and 1-1=0, those values are the range of your derivative as you notice 0 to 2 is always positive. This isn't calculus this is trig, perhaps you have forgotten some of your highschool math nothing to be ashamed of, you should brush up on some trig, math is deffinately accumulative.
  15. Apr 8, 2005 #14

    "....0 to 2 is always positive."

  16. Apr 8, 2005 #15


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    It may have been glossed over, but the function can be 0 at isolated points and still have an inverse. Look at x^3, for example. The only thing that happens is that the inverse isn't differentiable at these points. But it's still a valid function.
  17. Apr 8, 2005 #16

    "The only thing that happens is that the inverse isn't differentiable at these points. But it's (x^1/3) a valid function."

    True, but that's because the x^3 has an inflection point at x=0 rather than a local max or min. But the function in questioin, f(x) = x + cos(x), has lots of local maxs and mins. So what does that say about whether it has an inverse?
  18. Apr 8, 2005 #17


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    No it doesn't. To have a max or min, the derivative must change sign. Graph the function and you'll see, all those points where the derivative is zero are points of inflection.
  19. Apr 8, 2005 #18


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    If the function has a zero derivative anywhere, then the inverse (if it exists) will not be differentiable (the derivative would be infinity somewhere, which is really to say that it has no derivative at that point). Now, on some domain your function f has no zero derivatives, but if no domain is specified, it will have zero derivatives somewhere. BUT, that doesn't mean that it doesn't have an inverse, it just means that it doesn't have a differentiable inverse. To show that the function is invertible, you simply need to show that it is 1-1, which is precisely what the horizontal line test shows.

    What does it mean to be 1-1 (one to one)? It means that two different x values, x1 and x2, will not give the same value f(x1) = f(x2). And think about the horizontal line test, isn't that exactly what it's about? So, without a graph, can you show that there are no two different x values for which the function takes the same value? Suppose f(a) = f(b), i.e. f takes on the same value at a and at b.

    a + cos(a) = b + cos(b)
    [cos(b) - cos(a)]/(b - a) = -1

    This is to say that you can draw a secant through the curve (x, cos(x)) with slope -1, intersecting the curve at points (a, cos(a)) and (b, cos(b)). Suppose we have a pair of points through which the secant is less than -1. Then by mean value theorem, we'd have a place where cos(x) had a slope of less than -1, which is impossible, so -1 is the smallest possible value for a secant. Choose some c such that a < c < b, and we know that:

    [cos(c) - cos(a)]/(c - a) > -1

    Suppose it is greater than -1, then [cos(b) - cos(c)]/(b - c) < -1, which would imply there is some d in (c, b) with cos'(d) < -1, which is impossible, so it cannot be greater than -1. Suppose it is -1, then [cos(b) - cos(c)]/(b - c) = -1. Then there is some d in (c, b) with cos'(d) = -1, and some e in (a, c) with cos'(e) = -1. but this requires d - e = 2kpi, for some positive integer k. But then b - a > 2kpi, so:

    cos(b) - cos(a) < 2kpi, which is impossible because cos(b) - cos(a) < 2. We have a contradiction, so there are no a, b satisfying:

    [cos(b) - cos(a)]/(b - a) = -1

    hence our function is 1-1, and hence invertible.
  20. Apr 8, 2005 #19
    StatusX and what,

    You are right! Adding the x term is just enough keep f(x) from ever turning back down. I even graphed it on Xcel, but to get more cycles, I used sin(10x). The x term couldn't keep that one from going down, so I was sure I was right.

    Anyway, sorry about being so adamant. I'll try to be more careful.


    Edit: This also applies to AKG. Didn't see your post until mine was up.
    Last edited: Apr 8, 2005
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