Question

If $x+2|y|=3y$, where $y=f\left(x\right)$, then $f\left(x\right)$ is

• A. continuous everywhere
• B. differentiable everywhere
• C. discontinuous at $x=0$
• D. not differentiable anywhere

Answer 1

The correct option is A continuous everywhere

For $f\left(x\right)\ge 0$, $\left|f\left(x\right)\right|=f\left(x\right)$.
So,
$x+2\left|f\left(x\right)\right|=3f\left(x\right)\Rightarrow x+2f\left(x\right)=3f\left(x\right)\Rightarrow f\left(x\right)=x$.
But since this is for $f\left(x\right)\ge 0$, $f\left(x\right)=x\forall x\ge 0$.
For $f\left(x\right)<0$, $\left|f\left(x\right)\right|=-f\left(x\right)$.
So,
$x+2\left|f\left(x\right)\right|=3f\left(x\right)\Rightarrow x-2f\left(x\right)=3f\left(x\right)\Rightarrow f\left(x\right)=\frac{x}{5}$.
But since this is for $f\left(x\right)<0$, $f\left(x\right)=\frac{x}{5}\forall x<0$.
Checking for continuity at $x=0$:
$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\frac{x}{5}=0\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}x=0f\left(0\right)=0$
Since it is an algebraic function that changes its definition only at $x=0$, the function is continuous everywhere.
Checking for differentiability at $x=0$: the slope of the function changes at this point, so it is not differentiable here.
So, option A is the correct option