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Thread: Need help, stuck!

  1. #1
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    Need help, stuck!

    1. If g(x)= 3 + x + e^x, find g^-1(4).

    It seems as if I cannot isolate "x" by itself.

    2. lim x / ( √(1+3x) - 1 )
    x->0


    I know the answer is 2/3 (due to graphing), but I cannot prove this answer utilizing the Limit Laws.

    3. A Tibetan monk leaves the monastery at 7:00 A.M. and takes his usual path to the top of the mountain, arriving at 7:00 P.M. The following morning, he starts at 7:00 A.M. at the top and takes the same path back, arriving at the monastery at 7:00 P.M. Use the Intermediate Value Theorem to show that there is a point on the path that the monk will cross at exactly the same time of day on both days.

    I am utterly stumped on this problem. I realize that I have to chart the path on the same graph, but then what?

    4. Find lim x⇒->∞ f(x) if, for all x > 1,

    5√(x) / (√(x-1)) < f(x) < (10e^x - 21) / (2e^x)
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Mathn00b
    1. If g(x)= 3 + x + e^x, find g^-1(4).

    It seems as if I cannot isolate "x" by itself.
    You are being asked to find an $\displaystyle x$ such that:

    $\displaystyle
    4=3+x+e^x
    $,

    or:

    $\displaystyle
    1=x+e^x
    $

    Sketching/graphing the right hand side shows that it equals 1 near $\displaystyle x=0$, but $\displaystyle e^0=1$ exactly, so:

    $\displaystyle
    1=0+e^0
    $

    hence $\displaystyle g^{-1}(4)=0$

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Mathn00b
    2. lim x / ( √(1+3x) - 1 )
    x->0


    I know the answer is 2/3 (due to graphing), but I cannot prove this answer utilizing the Limit Laws.
    Here we use L'hopitals rule:

    If $\displaystyle \lim_{x\to a} f(a)=0$, and $\displaystyle \lim_{x\to a} g(a)=0$, then:

    $\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)}=\lim_{x \to a} \frac{f'(x)}{g'(x)}$

    In the case of your problem:

    $\displaystyle f(x)=x$, and $\displaystyle g(x)=\sqrt{1+3x}-1$, then:

    $\displaystyle f'(x)=1$,

    and:

    $\displaystyle g'(x)=\frac{3}{2\sqrt{1+3x}}$

    So:

    $\displaystyle
    \lim_{x \to 0} \frac{x}{\sqrt{1+3x}-1}=\lim_{x \to 0} \frac{1}{\frac{3}{2\sqrt{1+3x}}}=\frac{1}{\frac{3} {2}}=2/3
    $

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Mathn00b

    3. A Tibetan monk leaves the monastery at 7:00 A.M. and takes his usual path to the top of the mountain, arriving at 7:00 P.M. The following morning, he starts at 7:00 A.M. at the top and takes the same path back, arriving at the monastery at 7:00 P.M. Use the Intermediate Value Theorem to show that there is a point on the path that the monk will cross at exactly the same time of day on both days.

    I am utterly stumped on this problem. I realize that I have to chart the path on the same graph, but then what?
    Let $\displaystyle L$ be the length of the path, and let $\displaystyle d_1(t)$ be
    the distance that the monk is from the bottom at time $\displaystyle t$
    on day 1, and $\displaystyle d_2(t)$, be the distance the monk is form the bottom
    at time $\displaystyle t$ on day 2.

    Now consider the function: $\displaystyle f(t)=d_1(t)-d_2(t)$.

    At 7am $\displaystyle f(7am)=0-L=-L$,
    at 7pm $\displaystyle f(7pm)=L-0=L$.

    So by the intermediate value theorem there must be some time t between
    7am and 7pm where $\displaystyle f(t)=0$, as $\displaystyle 0$ is between $\displaystyle f(7am)$ and $\displaystyle f(7pm)$.

    ($\displaystyle f(t)=0$ of course means that at time $\displaystyle t$ on both days the monk is at
    the same disrance from the bottom, that is at the same point)

    RonL
    Last edited by CaptainBlack; Jul 8th 2006 at 11:08 PM.
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  5. #5
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    Hello, Mathn00b!

    $\displaystyle 2)\;\lim_{x\to0}\frac{x}{\sqrt{1+3x} - 1}$

    Raionalize the denominator; multiply top and bottom by $\displaystyle \sqrt{1+3x} + 1:$

    . . $\displaystyle \frac{x}{\sqrt{1+3x} - 1}\cdot\frac{\sqrt{1+3x} + 1}{\sqrt{1+3x} + 1} \;= $ $\displaystyle \;\frac{x(\sqrt{1+3x} + 1)}{(1+3x) - 1} \;=$ $\displaystyle \;\frac{x\sqrt{1+3x} + 1}{3x}\;=\;\frac{\sqrt{1+3x} + 1}{3}$


    Therefore: .$\displaystyle \lim_{x\to\0}\frac{\sqrt{1+3x} + 1}{3}\;= \;\frac{\sqrt{1} + 1}{3}\;=\;\frac{2}{3}$




    $\displaystyle 4)\text{ Find }\lim_{x\to\infty}f(x)\text{, if for all }x > 1:$ $\displaystyle \;\frac{5\sqrt{x}}{\sqrt{x-1}} \; < \;f(x) \;< \;\frac{10e^x - 21}{2e^x}$

    This is an application of the famous "Squeeze Theorem".


    Divide top and bottom of the first fraction by $\displaystyle \sqrt{x}:\;\;\frac{ \frac{5\sqrt{x}}{\sqrt{x}}}{ \frac{\sqrt{x-1}}{\sqrt{x}}} \;= $ $\displaystyle \;\frac{5}{\sqrt{1 - \frac{1}{x}}}$

    Divide top and bottom of the second fraction by $\displaystyle e^x:\;\;\frac{\frac{10e^x}{e^x} - \frac{21}{e^x}}{\frac{2e^x}{e^x}}\;=$ $\displaystyle \;\frac{10-\frac{21}{e^x}}{2} $


    We have: .$\displaystyle \lim_{x\to\infty}\frac{5}{\sqrt{1 - \frac{1}{x}}} \;< \:\lim_{x\to\infty} f(x) \;<\;\lim_{x\to\infty}\frac{10 - \frac{21}{e^x}}{2} $

    . . .Then: . . . .$\displaystyle \frac{5}{\sqrt{1 - 0}}\;<\;\lim_{x\to\infty}f(x)\;<\;\frac{10-0}{2}$

    . . Hence: . . . . . . . .$\displaystyle 5 \:< \:\lim_{x\to\infty}f(x) \:<\:5\quad \Leftarrow\;squeeze!$


    Therefore: .$\displaystyle \lim_{x\to\infty}f(x) \:=\:5$

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  6. #6
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    Thank you CaptainBlack and Soroban! I now understand my mistakes. Thanks for the extremely clear explanations!
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