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Math Help - 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 x such that:

    <br />
4=3+x+e^x<br />
,

    or:

    <br />
1=x+e^x<br />

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

    <br />
1=0+e^0<br />

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

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

    In the case of your problem:

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

    f'(x)=1,

    and:

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

    So:

    <br />
\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<br />

    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 L be the length of the path, and let d_1(t) be
    the distance that the monk is from the bottom at time t
    on day 1, and d_2(t), be the distance the monk is form the bottom
    at time t on day 2.

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

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

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

    ( f(t)=0 of course means that at time 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; July 8th 2006 at 11:08 PM.
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  5. #5
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    Hello, Mathn00b!

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

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

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


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




    4)\text{ Find }\lim_{x\to\infty}f(x)\text{, if for all }x > 1: \;\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 \sqrt{x}:\;\;\frac{ \frac{5\sqrt{x}}{\sqrt{x}}}{ \frac{\sqrt{x-1}}{\sqrt{x}}} \;= \;\frac{5}{\sqrt{1 - \frac{1}{x}}}

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


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

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


    Therefore: . \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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