Question

Discuss the continuity and differentiability of
$f\left(x\right)=$$\{\begin{array}{cc}2-x,& x<2\\ (2-x)(4-x),& 2\le x\le 4\\ 4-x,& x>4\end{array}$

Answer 1

When $x<2$, We have $f\left(x\right)=2-x$.
We know that, a polynomial function is continuous and differentiable everywhere. So, $f\left(x\right)$ is continuous and differentiable for all $x<2$.
Similarly, $f\left(x\right)$ is continuous and differentiable for all $x\in [2,4]$ and $x>4$.

Thus, the possible points where we have to check the continuity and differentiability of $f\left(x\right)$ are $x=2$ and $x=4$.

Continuity at $x=2$:
$\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{-}}{lim}2-x=2-2=0$
$\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}(2-x)(4-x)=0\times 2=0$
and, $f\left(2\right)=(2-2)(4-2)=0$
$\Rightarrow$ $\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}f\left(x\right)=f\left(2\right)$
So, $f\left(x\right)$ is continuous at $x=2$.

Continuity at $x=4$:
$\underset{x\to 4^{-}}{lim}f\left(x\right)=\underset{x\to 4^{-}}{lim}(2-x)(4-x)=(2-4)(4-4)=0$
$\underset{x\to 4^{+}}{lim}f\left(x\right)=\underset{x\to 4^{+}}{lim}(4-x)=(4-4)=0$
and, $f\left(4\right)=(2-4)(4-4)=0$
$\Rightarrow$ $\underset{x\to 4^{-}}{lim}f\left(x\right)=\underset{x\to 4^{+}}{lim}f\left(x\right)=f\left(4\right)$
So, $f\left(x\right)$ is continuous at $x=4$.

Differentiability at $x=2$:
$(\text{LHD at}x=2)=$$\underset{x\to 2^{-}}{lim}\frac{f\left(x\right)-f\left(2\right)}{x-2}=\underset{x\to 2^{-}}{lim}\frac{(2-x)-0}{x-2}=\underset{x\to 2^{-}}{lim}\frac{-(x-2)}{x-2}=-1$
$(\text{RHD at}x=2)=$ $\underset{x\to 2^{+}}{lim}\frac{f\left(x\right)-f\left(2\right)}{x-2}=\underset{x\to 2^{+}}{lim}\frac{(2-x)(4-x)-0}{x-2}=\underset{x\to 2^{+}}{lim}\frac{(x-2)(x-4)}{x-2}$
$=\underset{x\to 2^{+}}{lim}x-4=2-4=-2$

Thus, $(\text{LHD at}x=2)\ne (\text{RHD at}x=2)$
So, $f\left(x\right)$ is not differentiable at $x=2$.

Differentiability at $x=4$:
$(\text{LHD at}x=4)=$$\underset{x\to 4^{-}}{lim}\frac{f\left(x\right)-f\left(4\right)}{x-4}=\underset{x\to 4^{-}}{lim}\frac{(2-x)(4-x)-0}{x-4}=\underset{x\to 4^{-}}{lim}\frac{(x-2)(x-4)}{x-4}$
$=\underset{x\to 4^{-}}{lim}(x-2)=2$
$(\text{RHD at}x=4)=$$\underset{x\to 4^{+}}{lim}\frac{f\left(x\right)-f\left(4\right)}{x-4}=\underset{x\to 4^{+}}{lim}\frac{(4-x)-0}{x-4}=\underset{x\to 4^{+}}{lim}\frac{-(x-4)}{x-4}=-1$

Thus, $(\text{LHD at}x=4)\ne (\text{RHD at}x=4)$
$\Rightarrow f\left(x\right)$ is not differentiable at $x=4$.