If $f(x)=\left\{\begin{array}{ll}x^2(\operatorname{sgn}[x]+\{x\}), & 0

Question

If $f(x)=\left\{\begin{array}{ll}x^2(\operatorname{sgn}[x]+\{x\}), & 0<x<2 \ \sin x+|x-3|, & 2 \leq x<4\end{array}ight.$ where $\lfloor floor$ and $\{ \}$ represents the greatest integer and the fractional part function, respectively

Professor PC Thomas Classes · Answer

For continuity at $x=1$
$lim_{xightarrow1^{\prime}}f(x)=lim_{xightarrow1^{\prime}}(x^{2}sgn[x]+\{x\})=1+0=1$
$lim_{xightarrow1}f(x)=lim_{xightarrow1^{-1}}(x^{2}sgn[x]+\{x\})=1sgn(0)=1$
Also, $f(1)=1$
Hence $f(x)$ is continuous at $x=1$
Now for differentiability
$f^{\prime}(1^{+})=lim_{hightarrow0}\frac{(1+h)^{2}sgn[1+h]+\{1+h\}-1}{h} =lim_{hightarrow0}\frac{h^{2}+3h}{h}=3$
and $lf^{\prime}(-1)=lim_{hightarrow0}\frac{(1-h)^{2}sgn[1-h]+\{1-h\}-1}{h} =lim_{hightarrow0}\frac{h^{2}-3h}{-h}=3$
$f^{\prime}(1^{+})=f^{\prime}(1^{-})$
Hence, $f(x)$ is differentiable at $x=1$
Now at $x=2$
$lim_{xightarrow2^{-}}f(x)=lim_{xightarrow2^{-}}(x^{2}sgn[x]+\{x\})=4	imes0+1$
$lim_{xightarrow2^{+}}f(x)=lim_{hightarrow2^{+}}(sin~x+|x-3|)=1+sin~2$
Hence $LHL
e RHL$
Hence, $f(x)$ is discontinuous at $x=2$ and then $f(x)$ is also non differentiable at $x=2$