If f(x) = log(1 + x) + 1/(1+x), show that f(x) attains its minimum value at x = 0.

If $f(x) = \log(1 + x) + \frac{1}{1+x}$, show that $f(x)$ attains its minimum value at $x = 0$.

Given: $f(x) = \log(1 + x) + \frac{1}{1+x}$

To show: $f(x) \text{ attains its minimum value at } x = 0$

Answer

Proof that $f$ has its minimum at $x=0$ (domain $x>-1$)

1. $f'(x)=\frac1{1+x}-\frac1{(1+x)^2}$ (differentiating each term).
2. $f'(x)=\frac{(1+x)-1}{(1+x)^2}=\frac{x}{(1+x)^2}$.
3. For a stationary point $f'(x)=0$, so $x=0$ (as $(1+x)^2>0$ for $x>-1$).
4. $f''(x)=\frac{(1+x)^2-2x(1+x)}{(1+x)^4}=\frac{1-x}{(1+x)^3}$ (quotient rule).
5. $f''(0)=1>0$, so by the second derivative test $f$ has a minimum at $x=0$.
6. Hence $f(x)$ attains its minimum value at $x=0$, the minimum value being $f(0)=1$.