# Need help, stuck!

• July 8th 2006, 01:06 AM
Mathn00b
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)
• July 8th 2006, 02:15 AM
CaptainBlack
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:

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

or:

$
1=x+e^x
$

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

$
1=0+e^0
$

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

RonL
• July 8th 2006, 02:29 AM
CaptainBlack
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:

$
\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
• July 8th 2006, 02:43 AM
CaptainBlack
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
• July 8th 2006, 05:39 AM
Soroban
Hello, Mathn00b!

Quote:

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

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

Quote:

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

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

• July 8th 2006, 05:30 PM
Mathn00b
Thank you CaptainBlack and Soroban! I now understand my mistakes. Thanks for the extremely clear explanations! :)