Use derivatives to prove that if n\geq1, then (1+x)^{n}>1+nx for -1<x<0 and 0<x (notice that equality holds for x=0).

rocedwrp 2021-07-03 Answered

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

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Expert Answer

Brighton
Answered 2021-07-04 Author has 7775 answers

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

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