Prove with the mean value theorem that $x-\frac{x^2}{2}<\ln(1+x)
Approach
f(x):=ln(1+x) with the mean value theorem in [0,x]
11+ξ=ln(1+x)−0x−0
11+ξ takes the biggest value when ξ is 0
and so 1<ln(1+x)x multiply with x and you get
x<ln(1+x)
I can't prove the second part.
Answer
For MVT
ln(1+x)+x22−(ln1+0)=x⋅(11+c+c)>xc∈(0,x)
Indeed
11+c+c=c2+c+11+c>1
No comments:
Post a Comment