Use derivatives to prove that if n≥1, then (1+x)n&gt;1+nx for -1

Let n≥1. Then we have to show that (1+x){n}&gt;1+ns, for -10.Let us consider the function g(x)=(1+x)n—(1+nx). Then we note that g(0)=0. Moreover, by calculating the derivative we see thatg′(x)=n(1+x)n−1−n=n((1+x)n−1—1).Now since n−1≥1 we see that for x&gt;0,(1+x)n−1−1&gt;0,that is, g′(x)&gt;0 for x&gt;0. On the other hand, if -1 (1+x)n−1&lt;1which then shows that g′(x)&lt;0 on -10 and is decreasing on -10 on (0,∞) and g(x)&lt;0 on -1 (1+x)n&gt;1+nx, for -10.Thus, it completes the proof.Let n≥1. Then (1+x)n&gt;1+nx, for -10

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rocedwrp

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2021-07-03

Use derivatives to prove that if $n \geq 1$ , then ${(1 + x)}^{n} > 1 + n x$ for -1

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Brighton

Skilled2021-07-04Added 103 answers

Let $n \geq 1$ . Then we have to show that $(1 + x) {n} > 1 + n s$ , for -10.
Let us consider the function $g (x) = {(1 + x)}^{n} — (1 + n x)$ . Then we note that $g (0) = 0.$ Moreover, by calculating the derivative we see that
$g^{'} (x) = n {(1 + x)}^{n - 1} - n = n ({(1 + x)}^{n - 1} — 1)$ .
Now since $n - 1 \geq 1$ we see that for $x > 0$ ,
${(1 + x)}^{n - 1} - 1 > 0$ ,
that is, $g^{'} (x) > 0$ for $x > 0$ . On the other hand, if -1 ${(1 + x)}^{n - 1} < 1$
which then shows that $g^{'} (x) < 0$ on -10 and is decreasing on -10 on $(0, \infty)$ and $g (x) < 0$ on -1 ${(1 + x)}^{n} > 1 + n x$ , for -10.
Thus, it completes the proof.
Let $n \geq 1$ . Then $(1 + x)^{n} > 1 + n x,$ for -10

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