Answer
$1$
Work Step by Step
We are given the function:
$f(x)=e^x$
The average rate of change on $[x_0,x_1]$ is:
$\dfrac{\Delta f}{\Delta x}=\dfrac{f(x_1)-f(x_0)}{x_1-x_0}$
The instantaneous rate of change is the limit of the average rate of change.
In order to estimate the instantaneous rate of change at $x=0$, consider intervals $[x_1,x_0],[x_0,x_1]$ for $x_1$ close to $x_0$, where $x_0=0$:
$[-0.1,0]$: $\dfrac{\Delta f}{\Delta x}=\dfrac{f(-0.1)-f(0)}{-0.1-0}=\dfrac{e^{-0.1}-e^0}{-0.1}\approx 0.95162582$
See table
From the table we find that the instantaneous rate of of change at $x=0$ is approximately $1$.
