Explain why the expression x-(x+1)ln(x+1)<0 for positive values of x .
The derivative of the expression y = x - (x + 1)*ln(x +1) is
y' = 1 - ln(x + 1) - (x + 1)/(x + 1)
=> 1 - ln(x + 1) - 1
=> - ln(x + 1)
This shows that the function is decreasing for all positive values of x. Also, for x = 0 we have y = 0 - 1*ln 1 = 0
Therefore we can say that for all positive values of x the expression x - (x + 1)*ln(x + 1) < 0
We'll assign a function f(x)=x-(x+1)ln(x+1) and we'll have to prove that f(x)<0
To prove that the function has negative values, for any positive values of x, we'll must show that it's derivative is also negative.
We'll calculate it's derivative, using product rule:
f'(x) =1 - (x+1)' * ln(x+1) – (x+1)*[ln(x+1)]'
f'(x) = 1 - ln(x+1) - [(x+1)/(x+1)]*1
f'(x) = 1 - ln(x+1) - 1
We'll eliminate like terms:
f'(x) = -ln(x+1) < 0
The 1st derivative is negative.
Since x is positive => the function is negative, too.
Therefore, the inequality x-(x+1)ln(x+1)<0 is verified.