Question

$f\left(x\right)=\cos x$ and $g\left(x\right)=\{\begin{array}{cc}min\{f\left(t\right):0\le t\le x\},& 0\le x\le \pi \\ \sin x-1,& x>\pi \end{array}$.

Is $g\left(x\right)$ continuous in $(0,\infty )$?

Answer 1

Given $f\left(x\right)=\cos x$

$g\left(x\right)=\{\begin{array}{cc}min\{f\left(t\right):0\le t\le x\},& 0\le x\le \pi \\ \sin x-1,& x>\pi \end{array}$

Since, we know that $\cos x$ attains minimum value -1 at $x=\pi$ , when $x\in [0,\pi ]$

So, $g\left(x\right)=\{\begin{array}{cc}-1,& 0\le x\le \pi \\ \sin x-1,& x>\pi \end{array}$

Clearly, $g\left(x\right)=-1$ is continuous for $x\in [0,\pi$ , being a constant function.

Also, sine function is continuous everywhere. So, $g\left(x\right)=\sin x-1$ is continuous for all $x>\pi$

Now, we will check the continuity of $g\left(x\right)$ at $x=\pi$

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

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

$\Rightarrow LHL=-1$

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

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

$={lim}_{h\to 0}\sin (\pi +h)-1$

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

$=0-1$

$\Rightarrow RHL=-1$

Also, $f\left(\pi \right)=-1$

$\Rightarrow LHL=RHL=f\left(\pi \right)$

Hence, $f\left(x\right)$ is continuous at $x=\pi$