# Thread: Discuss continuity and differentiability of f(x)?

1. ## Discuss continuity and differentiability of f(x)?

A function f is defined by:
$f(x) = \arccos\left(Sgn\left(\frac{2[x]}{3x - [x]}\right)\right)$ where $Sgn(x) = \left\{ \begin{array}{cc}\frac{|x|}{x}, &\mbox{if}\ x\neq 0\\0, &\mbox{if}\ x = 0\end{array}\right.$ and [.] denotes the greatest integer function.

Discuss the continuity and differentiability of $f(x)$ at $x = \pm1$

2. Originally Posted by fardeen_gen
A function f is defined by:
$f(x) = \arccos\left(Sgn\left(\frac{2[x]}{3x - [x]}\right)\right)$ where $Sgn(x) = \left\{ \begin{array}{cc}\frac{|x|}{x}, &\mbox{if}\ x\neq 0\\0, &\mbox{if}\ x = 0\end{array}\right.$ and [.] denotes the greatest integer function.

Discuss the continuity and differentiability of $f(x)$ at $x = \pm1$
For continuity, I'm going to use the limit definition.

At $x=1$, we have:

$\lim_{x\to 1^-}f(x) = sgn\left(\frac{2(0)}{3x-0}\right) = 0. ~~~\arccos(0)=\frac{\pi}{2}$

$\lim_{x\to 1^+} f(x)= sgn\left(\frac{2(1)}{3x-1}\right) = sgn(2/2) = 1. ~~~\arccos(1) = 0$

$f(1) = sgn\left(\frac{2(1)}{3x-1}\right) = sgn(2/2) = 1. ~~~\arccos(1) = 0$

So $f(x)$ is not continuous at $x=1$.

At $x=-1$, we have:

$\lim_{x\to -1^-}f(x) = sgn\left(\frac{2(-2)}{3x-(-2)}\right) = sgn(-4/-1)=1. ~~~\arccos(1)=0$

$\lim_{x\to -1^+}f(x) = sgn\left(\frac{2(-1)}{3x-(-1)}\right) = sgn(-2/-2) = 1. ~~~\arccos(1) = 0$

$f(-1) = sgn\left(\frac{2(-1)}{3x-(-1)}\right) = sgn(-2/-2) = 1. ~~~\arccos(1) = 0$

So $f(x)$ is continuous at $x=-1$.

3. It follows that $f$ is not diffrentiable at $x=1$, because it is not continuous.

At $x=-1$, we have: $\lim_{h\to 0}\frac{sgn\left(\frac{2[-1+h]}{3(-1+h)-[-1+h]}\right)-sgn\left(\frac{2[-1]}{3(-1)-[-1]}\right)}{h} = \lim_{h\to 0}\frac{sgn\left(\frac{2[-1+h]}{3(-1+h)-[-1+h]}\right)-1}{h}$.

Looking at $\lim_{h\to 0^+}$ and assuming $0 (I think this is OK to assume because we want $h\to 0$), we know that $[h-1]=-1$ and $3h-2<0$, so as $h$ becomes less than $\frac{2}{3}$, we have:

$\lim_{h\to 0^+}\frac{sgn\left(\frac{-2}{3h-2}\right)-1}{h} = \lim_{h\to 0^+}\frac{sgn\left(\frac{-2}{<0}\right)-1}{h} = \lim_{h\to 0^+}\frac{1-1}{h} = 0$

Looking at $\lim_{h\to 0^-}$ and assuming $-1, we know that $[h-1]=-2$, and we have:

$\lim_{h\to 0^-}\frac{sgn\left(\frac{-4}{3h-1}\right)-1}{h} = \lim_{h\to 0^-}\frac{sgn\left(\frac{-4}{<0}\right)-1}{h} = \lim_{h\to 0^-}\frac{1-1}{h} = 0$

So I think $f(x)$ is differentiable at $x=-1$, and furthermore that $f'(-1)=0$ and the tangent line is $g(x)=1$.