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 December 18th, 2016, 04:01 AM #1 Newbie   Joined: Dec 2016 From: Sweden Posts: 13 Thanks: 1 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.
 December 18th, 2016, 04:38 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,370 Thanks: 1274 $f(x) = (1+x)^r - (1+r x)$ 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.
December 18th, 2016, 06:13 AM   #3
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 Originally Posted by romsek $f(x) = (1+x)^r - (1+r x)$ 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 $(1+x)^r > 1+r x,~x \in [-1,0) \cup (0, \infty)$
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