The set of all $x$ for which $\ln (1+x) \leq x$ is equal to ......

Question

Professor PC Thomas Classes · Answer

Let $f(x)=\log (1+x)-x$, for $x>-1$\ $f^{\prime}(x)=\frac{1}{1+x}-1=\frac{-x}{1+x}$\ $	herefore \quad f^{\prime}(x) \geq 0$, for all $x \in(-1,0)$\ and $f^{\prime}(x) \leq 0$, for all $x \geq 0$\ $	herefore \quad f$ is increasing for all $x \in(-1,0)$\ and $f$ is decreasing for all $x \geq 0$\ Also $f(0)=0$\ $f(x) \leq 0$, for all $x \in(-1,0)$\ and $f(x) \leq 0$, for all $x \geq 0$\ $\Rightarrow \quad \log (1+x)-x \leq 0$, for all $x>-1$\