Question

if x is positive show that x> log(1+x)> x-1/2x^2

Answer 1

SOLUTION

TO PROVE

$\displaystyle \sf{x > \log (1 + x) > x - \frac{ {x}^{2} }{2} \: \: \: when \: x > 0 }$

PROOF

First suppose that

$\displaystyle \sf{f(x) = x - \log (1 + x)}$

Then

$\displaystyle \sf{f'(x) = 1 - \frac{1}{1 + x} }$

$\displaystyle \sf{ \implies \: f'(x) = \frac{x}{1 + x} }$

$\displaystyle \sf{ \implies \: f'(x) > 0 \: \: as \: x > 0}$

So f(x) is increasing for x > 0

$\displaystyle \sf{ \implies \: f(x) > f(0) \: }$

$\displaystyle \sf{ \implies \: f(x) >0 \: \: \: as \: f(0) = 0}$

$\displaystyle \sf{\implies \: x - \log (1 + x) > 0}$

$\displaystyle \sf{\implies \: x > \log (1 + x) \: \: \: - - - (1)}$

Next suppose that

$\displaystyle \sf{g(x) = \log (1 + x) - \bigg( x - \frac{ {x}^{2} }{2} \bigg) \: \: \: when \: x > 0 }$

$\displaystyle \sf{ \implies \: g'(x) = \frac{1}{1 + x} - (1 - x) }$

$\displaystyle \sf{ \implies \: g'(x) = \frac{1}{1 + x} - 1 + x}$

$\displaystyle \sf{ \implies \: g'(x) = \frac{1 - 1 - x}{1 + x} + x}$

$\displaystyle \sf{ \implies \: g'(x) = x - \frac{ x}{1 + x} }$

$\displaystyle \sf{ \implies \: g'(x) = \frac{ {x}^{2} }{1 + x} }$

$\displaystyle \sf{ \implies \: g'(x) > 0 \: \: \: as \: x > 0 }$

So g(x) is increasing for x > 0

$\displaystyle \sf{ \implies \: g(x) > g(0) \: \: \: }$

$\displaystyle \sf{ \implies \: g(x) > 0\: \: \: as \: \: \: g(0) = 0}$

$\displaystyle \sf{ \implies \: \log (1 + x) - \bigg( x - \frac{ {x}^{2} }{2} \bigg) > 0\: \: \: when \: x > 0 }$

$\displaystyle \sf{ \implies \: \log (1 + x) > \bigg( x - \frac{ {x}^{2} }{2} \bigg) \: } \: \: - - - (2)$

Combining (1) & (2) we have

$\displaystyle \sf{x > \log (1 + x) > x - \frac{ {x}^{2} }{2} \: \: \: when \: x > 0 }$

Hence proved

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