The function f(x) = e^|x| is

The correct answer is Option (1) → continuous everywhere but not differentiable at $x = 0$ ##

Let $u(x) = |x|$ and $v(x) = e^x$

$∴f(x) = v \circ u(x) = v[u(x)] = v|x| = e^{|x|}$

Since, $u(x)$ and $v(x)$ are both continuous functions.

So, $f(x)$ is also continuous function but $u(x) = |x|$ is not differentiable at $x = 0$, whereas $v(x) = e^x$ is differentiable at everywhere.

Hence, $f(x)$ is continuous everywhere but not differentiable at $x = 0$.