Question

Examine the following curve for continuity and differentiability: $y=x^2$ for $x\le 0;y=1$ for$0x\le 1$ and$y=1/x$ for $x>1.$ Also draw the graph of the function.

Answer 1

$f\left(x\right)=\{\begin{array}{c}{x}^{2}x\le 0\\ 10<x\le 1\\ \frac{1}{x}x>1\end{array}$

$f\left(x\right)=x^2$ for $x<0$ is continuous, being a polynomial function.

$f\left(x\right)=1$ for $x\in (0,1)$ is continuous, being a constant function.

$f\left(x\right)=\frac{1}{x}$ for $x>1$ , again a part of reciprocal function, so continuous.

We need to check continuity at $x=0$ and $x=1$

At $x=0$

$LHL={lim}_{x\to 0^{-}}f\left(x\right)$

$={lim}_{h\to 0}f(0-h)$

$={lim}_{h\to 0}(-h{)}^2$

$\Rightarrow LHL=0$

$RHL={lim}_{x\to 0^{+}}f\left(x\right)$

$={lim}_{h\to 0}f(0+h)$

$\Rightarrow RHL=1$

$f\left(0\right)=1$

Since, $LHL\ne RHL$.

So, the function is discontinuous at $x=0$

At $x=1$

$LHL={lim}_{x\to 1^{-}}f\left(x\right)$

$={lim}_{h\to 0}f(1-h)$

$\Rightarrow LHL=1$

$RHL={lim}_{x\to 1^{+}}f\left(x\right)$

$={lim}_{h\to 0}f(1+h)$

$={lim}_{h\to 0}\frac{1}{1+h}$

$\Rightarrow RHL=1$

$f\left(1\right)=1$

Since, $LHL=RHL=f\left(\right)$.

So, the function is continuous at $x=1$

Since, $f\left(x\right)$ is not continuous at $x=0$

So, $f\left(x\right)$ is also not differentiable at $x=0$

Now, we will check differentiability at $x=1$

$L{f}^{\prime }\left(1\right)={lim}_{h\to 0}\frac{f(1-h)-f\left(1\right)}{-h}$

$={lim}_{h\to 0}\frac{1-1}{-h}$

$\Rightarrow L{f}^{\prime }\left(1\right)=0$

$R{f}^{\prime }\left(1\right)={lim}_{h\to 0}\frac{f(1+h)-f\left(1\right)}{h}$

$={lim}_{h\to 0}\frac{\frac{1}{1+h}-1}{h}$

$\Rightarrow R{f}^{\prime }\left(1\right)=0$

Since, $L{f}^{\prime }\left(1\right)=R{f}^{\prime }\left(1\right)$

Hence, $f\left(x\right)$ is differentiable at $x=1$