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December 18th, 2016, 04:01 AM   #1
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In one Variable; Using Mean Value Theorem to show inequality with another unknown

I can't understand the steps in this proof.

Using the Mean Value Theorem:

$\displaystyle Let \: r>1 \: . \: If \: x>0 \: or \: -1 \leq \: x \: <0. $

Show that $\displaystyle (1+x)^{r}>1+rx \: $

Please help me understand the steps in the proof.
I don't understand them;

$\displaystyle Let \: f(x)=(1+x)^{r}-1-rx \: where \: r>1. $
$\displaystyle Then \: f'(x)=(1+x)^{r-1}-r \: $
$\displaystyle If \: -1 \leq x < 0 \: then \: f'(x)<0; \: if \: x>0, \: then \: f'(x)>0. $
$\displaystyle Thus \: f(x)>f(0)=0 \: if \: -1 \leq x < 0 \: or \: x>0.$
$\displaystyle Thus \: (1+x)^{r}>1+rx \: if \: -1 \leq x < 0 \: or \: x>0.$

This is supposed to complete it.
But I do not understand a single line of it.

Stuck all summer with it.

Please help.

Last edited by Luciferis; December 18th, 2016 at 04:08 AM.
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December 18th, 2016, 04:38 AM   #2
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$f(x) = (1+x)^r - (1+r x)$

There should be no confusion about this line. They are simply defining $f$ as the difference of the terms involved in the inequality. Eventually I imagine it will be shown $f$ is non-negative.

$f^\prime(x) = r(1+x)^{r-1} - r$

This is a straightforward derivative. No reason for confusion here.

$-1 \leq x < 0 \Rightarrow f^\prime(x) < 0$

else

$0 \leq x \Rightarrow f^\prime(x) > 0$

The second of these should be pretty clear. For the first note that

$(1+x) < 1,~-1\leq x < 0 \Rightarrow (1+x)^{r-1} < 1

\Rightarrow r(1+x)^{r-1} - r = r((1+x)^{r-1}-1) < 0$

and also note that $f(0) = 1-1=0$

So for $0 < x$ we have that $f(x)$ increases from $0$ and thus $f(x)>0,~x>0$

For $-1\leq x < 0$ we have $f(x)$ decreases towards $0$

Well that must mean that it's decreasing from something greater than 0, so

$f(x) > 0,~-1 \leq x < 0$

and if $f(x) > 0, ~x\neq 0$ then $(1+x)^r > 1+r x,~x \in [-1,0) \cup (0, \infty)$

Last edited by romsek; December 18th, 2016 at 04:41 AM.
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December 18th, 2016, 06:13 AM   #3
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Quote:
Originally Posted by romsek View Post
$f(x) = (1+x)^r - (1+r x)$

There should be no confusion about this line. They are simply defining $f$ as the difference of the terms involved in the inequality. Eventually I imagine it will be shown $f$ is non-negative.

$f^\prime(x) = r(1+x)^{r-1} - r$

This is a straightforward derivative. No reason for confusion here.

$-1 \leq x < 0 \Rightarrow f^\prime(x) < 0$

else

$0 \leq x \Rightarrow f^\prime(x) > 0$

The second of these should be pretty clear. For the first note that

$(1+x) < 1,~-1\leq x < 0 \Rightarrow (1+x)^{r-1} < 1

\Rightarrow r(1+x)^{r-1} - r = r((1+x)^{r-1}-1) < 0$

and also note that $f(0) = 1-1=0$

So for $0 < x$ we have that $f(x)$ increases from $0$ and thus $f(x)>0,~x>0$

For $-1\leq x < 0$ we have $f(x)$ decreases towards $0$

Well that must mean that it's decreasing from something greater than 0, so

$f(x) > 0,~-1 \leq x < 0$

and if $f(x) > 0, ~x\neq 0$ then $(1+x)^r > 1+r x,~x \in [-1,0) \cup (0, \infty)$
Flawless 😀

Thank you.

Merry Christmas!!!!
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