Question

Find continuously of $f\left(x\right),$ if $f\left(x\right)=\left\{\begin{array}{c}x+1ifx\ge 1\\ x^2+1ifx<1\end{array}\right\}$

Answer 1

$f\left(x\right)=\left\{\begin{array}{c}x+1ifx\ge 1\\ x^2+1ifx<1\end{array}\right\}$

Since f(x) is polynomial functions, Hence f(x) is

continuous at $x>1$ and $x<1$

The only point we need to check at x =1

$a+x=1,f\left(x\right)=x+1\Rightarrow f\left(1\right)=1+1=2$

Now $\underset{x\to 1}{lim}f\left(x\right)$

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{n\to 0}{lim}f(1+h)=\underset{n\to 0}{lim}f(1+h)+1=2.$

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{n\to 0}{lim}f(1-h)=\underset{n\to 0}{lim}f(1-h^2)+1=\underset{n\to 0}{lim}(1+h^2-2h+1)=2$

$\therefore \underset{x\to 1^{+}}{lim}f\left(x\right)=2=\underset{x\to 1}{lim}f\left(x\right)\Rightarrow \underset{x\to 1}{lim}f\left(x\right)=2.$

$\therefore \underset{x\to 1}{lim}f\left(x\right)=f\left(1\right)$

$\therefore f\left(x\right)$ is continuous at x = 1

$\therefore f\left(x\right)$ is continuous everywhere.