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Math Help - This two problems are killing me

  1. #1
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    This two problems are killing me

    My body ask me for help, so Could any of you solve this for me so that I can to explain. It would be greatly appreciate. Thanks
    Attached Thumbnails Attached Thumbnails This two problems are killing me-20140501_213116.jpg  
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  2. #2
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    Re: This two problems are killing me

    for the first problem just differentiate both sides with respect to $x$ and solve the result (simple) differential equation for $f(x)$
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    Re: This two problems are killing me

    Quote Originally Posted by Pardovani View Post
    My body ask me for help, so Could any of you solve this for me so that I can to explain. It would be greatly appreciate. Thanks
    for the 2nd one do you mean the limit as $x \to a$ ?

    The limit as $x \to 0$ is just $\dfrac {f(0)-f(a)}{\sqrt{a}}$
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    Re: This two problems are killing me

    Yes, and what do you mean about the first problems, I don't see how I can solve it. Could you please show me some steps process thanks.
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    Re: This two problems are killing me

    Quote Originally Posted by Pardovani View Post
    Yes, and what do you mean about the first problems, I don't see how I can solve it. Could you please show me some steps process thanks.
    can you take the derivative with respect to x of both sides?
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  6. #6
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    Re: This two problems are killing me

    for the first problem I dont know how to do it
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    Re: This two problems are killing me

    Quote Originally Posted by Pardovani View Post
    for the first problem I dont know how to do it
    surely you can take the derivative of the right hand side?

    for the left hand side you use the Liebniz rule

    $\dfrac {d}{dx} \displaystyle{\int_{\ell(x)}^{u(x)}}f(t)~dt = f(u(x))\dfrac {du}{dx}-f(\ell(x))\dfrac{d\ell}{dx}$
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  8. #8
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    Re: This two problems are killing me

    Quote Originally Posted by romsek View Post
    for the 2nd one do you mean the limit as $x \to a$ ?

    The limit as $x \to 0$ is just $\dfrac {f(0)-f(a)}{\sqrt{a}}$
    this one is a little tricky but not that bad.

    $\displaystyle{\lim_{x\to a}}~\dfrac {f(x)-f(a)}{\sqrt{x}-\sqrt{a}}=\displaystyle{\lim_{x\to a}}\dfrac {f(x)-f(a)}{\sqrt{x}-\sqrt{a}}\dfrac {\sqrt{x}+\sqrt{a}} {\sqrt{x}+\sqrt{a}}=$

    $\displaystyle{\lim_{x\to a}}~\dfrac {-\sqrt{x} f(a)-\sqrt{a} f(x)+\sqrt{a} f(a)+\sqrt{x} f(x)}{x-a}$

    now use L'Hopital's rule

    $n(x)=-\sqrt{x} f(a)-\sqrt{a} f(x)+\sqrt{a} f(a)+\sqrt{x} f(x)$

    $d(x) = x-a$

    $\dfrac {dn}{dx}=-\sqrt{a} f'(x)-\frac{f(a)}{2 \sqrt{x}}+\sqrt{x} f'(x)+\frac{f(x)}{2 \sqrt{x}}$

    $\dfrac {dd}{dx}=1$

    $\displaystyle{\lim_{x\to a}}~-\sqrt{a} f'(x)-\frac{f(a)}{2 \sqrt{x}}+\sqrt{x} f'(x)+\frac{f(x)}{2 \sqrt{x}}=$

    $\displaystyle{\lim_{x\to a}}~\dfrac{f(x)-f(a)}{2\sqrt{x}}+f^\prime(x)\left(\sqrt{x}+\sqrt{a }\right)=2 f^\prime(a) \sqrt{a}$ so

    $\displaystyle{\lim_{x\to a}}~\dfrac {f(x)-f(a)}{\sqrt{x}-\sqrt{a}}=2 f^\prime(a) \sqrt{a}$

    I wouldn't be shocked if there's a much cleverer way to do this.
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  9. #9
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    Re: This two problems are killing me

    Great I really appreciate, omg it was so simple. But thanks
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    Re: This two problems are killing me

    So for the first problem, Do we just have to find different formula that will equal to [f(x^2)], there are no number and nothing so I wasn't really sure about what will be my first 2 steps
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  11. #11
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    Re: This two problems are killing me

    I see your formula but what is that going to lead us to?
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  12. #12
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    Re: This two problems are killing me

    Quote Originally Posted by Pardovani View Post
    I see your formula but what is that going to lead us to?
    If you are referring to the Liebniz formula in post 7, u(x)= x, so u'= 1, and w(x)= 0, so w'= 0.
    \frac{d}{dx}\int_0^x f(t)dt= f(x)
    (That's really just the "Fundamental Theorem of Calculus", you don't need the full "Liebniz formula".)
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    Re: This two problems are killing me

    Ok thanks, but it's fine I still dont really know what to do to solve it. I have a test soon and my professor was saying that it could be on it. but again thanks.
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  14. #14
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    Re: This two problems are killing me

    Quote Originally Posted by Pardovani View Post
    Ok thanks, but it's fine I still dont really know what to do to solve it. I have a test soon and my professor was saying that it could be on it. but again thanks.
    As Halls noted the derivative of left side is simply $f(x)$ You really should be able to see that immediately from a first glance.

    $\dfrac d {dx} \displaystyle{\int_0^x} f(t)~dt = \dfrac d {dx} \left(f(x)\right)^2$

    $f(x) = 2 f(x) f^\prime(x)$

    This is of course satisfied if $f(x)=0$ otherwise

    $f^\prime(x)=\dfrac 1 2$

    $f(x)=\dfrac x 2 + C$

    checking we find

    $\dfrac {x^2}{4} + Cx \overset{?}{=} \dfrac{x^2}{4} + 2Cx + C^2$

    and thus it's clear that $C=0$ so

    $f(x)=0 \mbox{ or }f(x)=\dfrac x 2$
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  15. #15
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    Re: This two problems are killing me

    Hi,
    Romsek's conclusion about your first problem is correct, but I believe his proof is incomplete. The problem is that you need the derivative f' to be 1/2 for all x; I don't think this is obvious. Here's my solution:

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