Question

The function $f\left(x\right)=\frac{x}{1+\left|x\right|}$ is differentiable on

• A. $(0,\infty )$
• B. $[0,\infty )$
• C. $(-\infty ,0)$
• D. $(-\infty ,\infty )$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $(0,\infty )$
B $[0,\infty )$
C $(-\infty ,0)$
D $(-\infty ,\infty )$

The given function can be written as
$f\left(x\right)=\{\begin{array}{cc}\frac{x}{1+x}& \text{if}x\ge 0\\ \frac{x}{1-x}& \text{if}x<0\end{array}$
Since $x/(1+x)(x>0)$ and $x/(1-x)(x<0)$ have non-zero polynomials in their denominators, they are differentiable in their respective domains. For $x=0$, we check directly that
$f^{\prime }(0+)=\underset{h\to 0+}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}=\underset{h\to 0+}{lim}\frac{h/(1+h)-0}{h-0}$
$=\underset{h\to 0+}{lim}\frac{1}{1+h}=1$
$f^{\prime }(0-)=\underset{h\to 0-}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}=\underset{h\to 0-}{lim}\frac{h/(1-h)-0}{h-0}$
$=\underset{h\to 0-}{lim}\frac{1}{1-h}=1$
Thus f is derivable at $x=0$ also, and hence f is derivable everywhere.