Question

If $x+|y|=2y,$ then $y$ as a function of $x$ is
$(\text{where}y=f(x\left)\right)$

• A. defined for all real $x$
• B. continuous at $x=0$
• C. differentiable at $x=0$
• D. $f^{\prime }\left(0^{-}\right)=\frac{1}{3}$

Answer 1

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

$f^{\prime }\left(0^{-}\right)=\frac{1}{3}$

Given : $x+|y|=2y$
$\Rightarrow y=f\left(x\right)=\{\begin{array}{cc}x& \text{if}y\ge 0,x\ge 0\\ \frac{x}{3}& \text{if}y<0,x<0\end{array}$
$y$ as a function of $x$ is defined for all $x\in R$
As $f\left(0^{+}\right)=f\left(0^{-}\right)=f\left(0\right)$
Hence, $f\left(x\right)$ is a continuous at $x=0.$

$f^{\prime }\left(0^{+}\right)=\underset{h\to 0^{+}}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0^{+}}{lim}\frac{h}{h}=1f^{\prime }\left(0^{-}\right)=\underset{h\to 0^{+}}{lim}\frac{f(0-h)-f\left(0\right)}{-h}=\underset{h\to 0^{+}}{lim}\frac{-\frac{h}{3}}{-h}=\frac{1}{3}\therefore R.H.D.\ne L.H.D.$
Hence, $f\left(x\right)$ is not differentiable at $x=0.$