5. Use the sign of the derivative to establish the inequaltiy ln(1 + x) > x − x 2 2 , ∀ x > 0
To establish this inequality we have to made a assumption first.
f(x)=ln(1+x)−x+x2f(x) = ln(1+x) -x+x^2f(x)=ln(1+x)−x+x2
Hence,
f′(x)=11+x−1+2xf'(x ) = \dfrac{1}{1+x}-1+2xf′(x)=1+x1−1+2x
f′(x)=x+2x21+xf'(x) = \dfrac{x+2x^2}{1+x}f′(x)=1+xx+2x2
For x>0x>0x>0 , We get f′(x)>0f'(x) >0f′(x)>0
Hence, we clearly say that f(x)>f(0)f(x)>f(0)f(x)>f(0)
Hence, ln(1+x)−x+x2>0ln(1+x)-x+x^2>0ln(1+x)−x+x2>0
OR
ln(1+x)>x−x2ln(1+x)>x-x^2ln(1+x)>x−x2 (Proved).
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