if u want to think abt it intuitively then:
\text{Consider}\ f(x)=x-\ln(1+x)
\text{Then},\ f^{\prime}(x)=1-\frac{1}{1+x}
\text{Let} \ f^{\prime}(x)=0
1-\frac{1}{1+x}=0
\frac{1}{1+x}=1
1+x=1
x=0
Therefore, turning point at x=0.
f^{\prime\prime}(x)=\frac{1}{(x+1)^2}\geq0 \ \text{for all} \ x > 0...