Question

Let two functions are defined as $g\left(x\right)=\{\begin{array}{cc}{x}^2,& -1\le x\le 2\\ x+2,& 2\le x\le 3\end{array},$ and $f\left(x\right)=\{\begin{array}{cc}x+1& x\le 1\\ 2x+1& 1<x\le 2,\end{array}$ then find $gof$(1)

Answer 1

From the given functions we get
$f\left(1\right)=2$ since $f\left(x\right)=x+1$ for all $x\le 1$...(i)
Now
$\underset{x\to 2^{-}}{lim}g\left(x\right)$
$=\underset{x\to 2^{-}}{lim}\left(x^2\right)$
$=4$
$\underset{x\to 2^{+}}{lim}g\left(x\right)$
$=\underset{x\to 2^{+}}{lim}(x+2)$
$=4$
Hence
$LHL=RHL$.
Thus the function $g\left(x\right)$ is continuous at $x=2$.
Now
$g\left(f\right(1\left)\right)$
$=g\left(2\right)$ since $f\left(1\right)=2$
$=4$.... from above limit calculation.
Hence
$g\left(f\right(1\left)\right)=4$