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Q. Consider the functions $f_{1}(x)=x, f_{2}(x)=2+\log _{e} x, x>0$. The graphs of the functions intersect.

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Solution:

$f_{1}(x)=x, f_{2}(x)=2+\log _{e} x$
Let $g(x)=f_{2}(x)-f_{1}(x)=2+\log _{e} x-x$
$g\left(0^{+}\right)<0, g(1)>0, g(e)>0, g\left(e^{2}\right)<0$
and value of $g(x)$ for all $x \geq e^{2}$ is negative.
$\therefore g(x)=0$ has two roots in $(0,1)$ and $\left(e, e^{2}\right)$