The function f(x) = e∣ˣ∣ is

The correct option is: "continuous everywhere but not differentiable at x = 0".

To understand why this is correct, let us analyze the continuity and differentiability of the function:
$$f(x)=e^{|x|}$$
Recall that the absolute value function is defined piecewise as:
$$|x|=\{\begin{array}{ll}x& \text{if}x\ge 0\\ -x& \text{if}x<0\end{array}$$
Therefore, we can rewrite the function $f(x)$ piecewise as:
$$f(x)=\{\begin{array}{ll}{e}^x& \text{if}x\ge 0\\ e^{-x}& \text{if}x<0\end{array}$$

Step 1: Check Continuity everywhere
For $x>0$, $f(x)=e^x$, which is an exponential function and is continuous.
For $x<0$, $f(x)=e^{-x}$, which is also an exponential function and is continuous.
Now, we check the continuity at the transition point $x=0$:
1. Left-Hand Limit (LHL) as $x\to 0^{-}$:
$$\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{-}}{lim}{e}^{-x}=e^0=1$$
2. Right-Hand Limit (RHL) as $x\to 0^{+}$:
$$\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}{e}^x=e^0=1$$
3. Function value at $x=0$:
$$f(0)=e^{|0|}=e^0=1$$
Since $\text{LHL}=\text{RHL}=f(0)=1$, the function $f(x)$ is continuous at $x=0$. Thus, $f(x)$ is continuous everywhere.

Step 2: Check Differentiability at x = 0
To determine differentiability at $x=0$, we calculate the Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD) at this point:
1. Left-Hand Derivative (LHD) at $x=0$:
$$\text{LHD}=\underset{h\to 0^{-}}{lim}\frac{f(0+h)-f(0)}{h}=\underset{h\to 0^{-}}{lim}\frac{e^{-h}-1}{h}$$
Using the standard limit $\underset{u\to 0}{lim}\frac{e^u-1}{u}=1$, let $u=-h$:
$$\text{LHD}=\underset{h\to 0^{-}}{lim}-\left(\frac{e^{-h}-1}{-h}\right)=-(1)=-1$$
2. Right-Hand Derivative (RHD) at $x=0$:
$$\text{RHD}=\underset{h\to 0^{+}}{lim}\frac{f(0+h)-f(0)}{h}=\underset{h\to 0^{+}}{lim}\frac{e^h-1}{h}=1$$
Since $\text{LHD}=-1$ and $\text{RHD}=1$, we have:
$$\text{LHD}\ne \text{RHD}$$
Thus, the function is not differentiable at $x=0$ (geometrically, the graph of $e^{|x|}$ has a sharp corner or "V-shape" at $x=0$).

Conclusion
The function $f(x)=e^{|x|}$ is continuous everywhere but is not differentiable at $x=0$.