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Math Help - Monotonically Increasing Proof

  1. #1
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    Monotonically Increasing Proof

    Suppose the function
    f has the following four properties:
    1.
    f is continuous for x >=0;

    2.
    f'(x) exists for x > 0;

    3.
    f(0) = 0;

    4.
    f'is monotonically increasing.
    Put

    g
    (x) =f(x)/x

    ;
    (x > 0)
    and prove that
    g is monotonically increasing. To better understand the various hypotheses,

    sketch the graph of a couple of typical functions
    f and interpret g geometrically.


    I started by trying some various functions for f.
    I tried f(x)=x,f(x)=x^1/2. Those are the two I looked at so far. I was trying to see if maybe there were better ones to sketch.
    Now interpreting g geometrically has me a little confused. I guess that would mean just sketching g. Now if f(x)=x, then g is the graph of 1. That doesn't seem right for the g to be monotonically increasing. For the othr, we have g=x^1/2/x=x^-1/2. Again, I get the feeling that g is not monotonically increasing. So, maybe I should have used different functions for f?
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  2. #2
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    Quote Originally Posted by mathematic View Post
    Suppose the function f has the following four properties:
    1. f is continuous for x >=0;
    2. f'(x) exists for x > 0;
    3. f(0) = 0;
    4. f' is monotonically increasing.
    Put g(x) = f(x)/x; (x > 0)
    and prove that g is monotonically increasing. To better understand the various hypotheses, sketch the graph of a couple of typical functions f and interpret g geometrically.

    I started by trying some various functions for f. I tried f(x)=x, f(x)=x^1/2. Those are the two I looked at so far. I was trying to see if maybe there were better ones to sketch

    Now interpreting g geometrically has me a little confused. I guess that would mean just sketching g. Now if f(x)=x, then g is the graph of 1. That doesn't seem right for the g to be monotonically increasing. For the other, we have g=x^1/2/x=x^-1/2. Again, I get the feeling that g is not monotonically increasing. So, maybe I should have used different functions for f?
    Condition 4. says that f' (the derivative of  f) must be monotonically increasing. That condition is not satisfied by either of your two trial functions. So yes, you should have used different functions for  f.
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  3. #3
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    Would f(x)=x^2 and f(x)=x^3 work? Both of those are increasing proved x>=0.
    Now f'(x)=2x and f'(x)=3x^2. Increasing when x>0
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathematic View Post
    Suppose the function
    f has the following four properties:
    1.
    f is continuous for x >=0;

    2.
    f'(x) exists for x > 0;

    3.
    f(0) = 0;

    4.
    f'is monotonically increasing.
    Put

    g
    (x) =f(x)/x

    ;
    (x > 0)
    and prove that
    g is monotonically increasing. To better understand the various hypotheses,

    sketch the graph of a couple of typical functions
    f and interpret g geometrically.


    I started by trying some various functions for f.
    I tried f(x)=x,f(x)=x^1/2. Those are the two I looked at so far. I was trying to see if maybe there were better ones to sketch.
    Now interpreting g geometrically has me a little confused. I guess that would mean just sketching g. Now if f(x)=x, then g is the graph of 1. That doesn't seem right for the g to be monotonically increasing. For the othr, we have g=x^1/2/x=x^-1/2. Again, I get the feeling that g is not monotonically increasing. So, maybe I should have used different functions for f?
    Merely note that if one let's \displaystyle g(x)=\frac{f(x)}{x} then one has \displaystyle g'(x)=\frac{f'(x)x-f(x)}{x^2} so it suffices to prove that f'(x)x\geqslant f(x) for all x>0. To see this let x_0>0 be fixed. Then, by assumption we may apply the MVT on [0,x_0] to get that \displaystyle \frac{f(x_0)}{x_0}=\frac{f(x_0)-f(0)}{x_.-0}=f'(\xi) for some \xi\in(0,x_0) but since f' is increasing we may conclude that \displaystyle f'(x_0)\geqslant f'(\xi)=\frac{f(x_0)}{x_0}. Since x_0 was arbitrary the conclusion follows.
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  5. #5
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    Well, I'm supposed to show this without the MVT. That theorem doesn't come till the next section in the book.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathematic View Post
    Well, I'm supposed to show this without the MVT. That theorem doesn't come till the next section in the book.
    Ok, what machinery do you have?
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  7. #7
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    Ok, the chapter I'm working out of is on the derivative. In particular, the last section we did was on derivatives and the intermediate value property. I have the limit definition of a derivative, definition of differentiable, combinations of differentiable functions(product rule, quotient rule, etc), chain rule, Darboux's Theorem, Interior Extremum Theorem.
    Interior Extremum Theorem says:Let f be differentiable on (a,b). If f attains a maximum value at some point c in (a,b) then f'(c)=0.
    Darbooux's Theorem says:If f is differentiable on an interval [a,b] and if d satisfies f'(a)<d<f'(b) then there exists a point c in (a,b) where f'(c)=d.

    Previous chapters have been on sequences, series, open/closed/compact sets, limits and continuity.
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  8. #8
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    Does any of that work? If none of what I have so far will be helpful, I will use the MVT, but I was iffy on using that since we only talked about that in the last 5 min of class.
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  9. #9
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    Are you sure this comes before MVT? This exact problem is in Baby Rudin and it comes after MVT there. But if you really can't use MVT then I can't help you because my solution is the same as Drexel's
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  10. #10
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    Ok well I could use the mvt. This wasn't assigned from our book, but all other problems assigned were out of the previous section, so I just assumed that this one I would use the previous section.
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