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Thread: Show that sqrt(1+x) <1+(1/2)x if x>0 ??

  1. #1
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    Show that sqrt(1+x) <1+(1/2)x if x>0 ??

    I have another test question i got wrong and didnt undertand soo im hoping someone can help me out.

    Show that sqrt(1+x) <1+(1/2)x if x>0 or ( formally written in attachment)
    Show that sqrt(1+x) &lt;1+(1/2)x if x&gt;0 ??-msp160019bb47e422a3f76500004b16f392eaba4c0f.gif

    I used the formula f'(c)= [f(b)-f(a)]/(b-a) and then just ended up with a mess at the end which is probably why i got it wrong. How would i go abouts starting this problem?
    Last edited by kensington; Jul 14th 2010 at 12:26 PM.
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  2. #2
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    Just square both sides

    $\displaystyle 4+4x<4+x^2+4x$

    $\displaystyle x^2>0$
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  3. #3
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    Quote Originally Posted by fobos3 View Post
    Just square both sides

    $\displaystyle 4+4x<4+x^2+4x$

    $\displaystyle x^2>0$

    Im a tad confused where the 4's came from.
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  4. #4
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    Multiply both side by 2. $\displaystyle 2\sqrt{1+x}<x+2$
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  5. #5
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    Quote Originally Posted by kensington View Post
    I have another test question i got wrong and didnt undertand soo im hoping someone can help me out.

    Show that sqrt(1+x) <1+(1/2)x if x>0 or ( formally written in attachment)
    Click image for larger version. 

Name:	MSP160019bb47e422a3f76500004b16f392eaba4c0f.gif 
Views:	105 
Size:	890 Bytes 
ID:	18197

    I used the formula f'(c)= [f(b)-f(a)]/(b-a) and then just ended up with a mess at the end which is probably why i got it wrong. How would i go abouts starting this problem?
    If x is positive, prove

    $\displaystyle \sqrt{1+x}<1+\frac{1}{2}x$

    $\displaystyle 1+\frac{1}{2}x=\frac{2+x}{2}$

    which is the positive square root of $\displaystyle \frac{4+4x+x^2}{4}$

    Taking it for granted that we are referring to the positive square root of 1+x, then

    $\displaystyle \sqrt{1+x}<\sqrt{\frac{x^2+4x+4}{4}}\ ?$

    $\displaystyle \sqrt{\frac{4x+4}{4}}<\sqrt{\frac{x^2+4x+4}{4}}\ ?$

    Yes
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  6. #6
    Senior Member roninpro's Avatar
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    You can also use calculus to prove the inequality.

    Consider the function $\displaystyle g(x)=\sqrt{x}-(1+\frac{1}{2}x)$. It suffices to show that when $\displaystyle x>0$, we have $\displaystyle g(x)>0$. First check $\displaystyle g(0)=0$ and find $\displaystyle g'(x)$ and show that it is always positive. Then you can conclude.
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  7. #7
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    You'd need a variation on that.
    If you prove a derivative is always positive,
    then you are proving the function is always increasing.

    Instead....

    $\displaystyle g(x)=(1+x)^{0.5}-1-0.5x$

    Locate the maxmum value of g(x)

    If $\displaystyle g'(x)=0\ for\ x=k$

    then if g(k)<0 the inequality is proven.



    Alternatively

    $\displaystyle h(x)=1+0.5x-(1+x)^{0.5}$

    Find the value of x=k giving minimum h(x),
    then show h(k)>0

    $\displaystyle 0.5-0.5(1+x)^{-0.5}=0$

    $\displaystyle \frac{1}{\sqrt{1+x}}=1$

    x=0.

    2nd derivative test at x=0.....$\displaystyle 0.25(1+x)^{-1.5}$

    this is 0.25 at x=0, indicating a minimum on the curve.

    h(0)=0

    This means that if x is positive, h(x)>0
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  8. #8
    Senior Member roninpro's Avatar
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    Quote Originally Posted by Archie Meade View Post
    You'd need a variation on that.
    If you prove a derivative is always positive,
    then you are proving the function is always increasing.
    We also require $\displaystyle g(0)=0$, which I mentioned in my post.

    For a more formal conclusion: we have $\displaystyle g(0)=0$ and $\displaystyle g(x)>0$ for all $\displaystyle x>0$. Then, by the Mean Value Theorem, there is $\displaystyle c\in (0, x)$ such that $\displaystyle \frac{g(x)-g(0)}{x-0}=g'(c)$. But this simplifies to $\displaystyle \frac{g(x)}{x}=g'(c)>0$. Multiplying $\displaystyle x$ to both sides gives the desired result.
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    Quote Originally Posted by roninpro View Post
    We also require $\displaystyle g(0)=0$, which I mentioned in my post.

    For a more formal conclusion: we have $\displaystyle g(0)=0$ and $\displaystyle g(x)>0$ for all $\displaystyle x>0$. Then, by the Mean Value Theorem, there is $\displaystyle c\in (0, x)$ such that $\displaystyle \frac{g(x)-g(0)}{x-0}=g'(c)$. But this simplifies to $\displaystyle \frac{g(x)}{x}=g'(c)>0$. Multiplying $\displaystyle x$ to both sides gives the desired result.
    Thats the formula i mentioned before but this whole g(x) thing is confusing me
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  10. #10
    Senior Member roninpro's Avatar
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    What part of $\displaystyle g(x)$ is causing you trouble?
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  11. #11
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    Quote Originally Posted by roninpro View Post
    What part of $\displaystyle g(x)$ is causing you trouble?
    So im looking at my original question with the square roots and inequality and i dont see how i can take those values and sub then in the mean value Theorem.
    I basically wote it as [1+(1/2)x - sqrt(1+x)]/ [(1/2)-(1+x)^(1/2)]... then eaither i got that wrong or i messed something up in simplifing because i couldnt prove thats its >0
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  12. #12
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    Do you want to analyse that with the mean value theorem?

    An algebraic or graphical solution is quite basic..

    $\displaystyle f(x)=\sqrt{1+x}$

    $\displaystyle g(x)=\frac{2+x}{2}$

    $\displaystyle \sqrt{1+x}=\sqrt{\frac{4+4x}{4}}$

    $\displaystyle \sqrt{\frac{4x+4}{4}}=\sqrt{\frac{x^2+4x+4}{4}}$

    only when x=0.

    Hence if x is positive, and since both functions are increasing functions

    $\displaystyle f'(x)=\frac{1}{2\sqrt{1+x}}$

    $\displaystyle g'(x)=\frac{1}{2}$

    For x>0, g'(x)>f'(x) which means that after x=0, g(x) is increasing more than f(x)
    for all x>0.

    Show that sqrt(1+x) &lt;1+(1/2)x if x&gt;0 ??-increasing-functions.jpg
    Last edited by Archie Meade; Jul 17th 2010 at 01:08 PM.
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