Answer
$f(0.1)f(1)<0,f$ continuous$\Rightarrow \exists x_0\in(0.1,1),f(x_0)=0$
Work Step by Step
We are given the function:
$f(x)=e^x+\ln x$
First we determine the domain of $f(x)$.
The domain of $e^x$ is $(-\infty,\infty)$, while the domain of $\ln x$ is $(0,\infty)$,therefore the domain of $f$ is:
$(-\infty,\infty)\cap (0,\infty)=(0,\infty)$
Choose two points in the domain and compute the value of the function in them:
$f(0.1)=e^{0.1}+\ln 0.1\approx -1.197$
$f(1)=e^{1}+\ln 1\approx 2.718$
As $f(x)$ is continuous on $(0.1,1)$ and $f(0.1)$ and $f(1)$ have opposite signs, it means that the function has a zero between 0.1 and 1.
We can tell that there is a zero between $0.1$ and $1$.
