The function $f(x) = {e^{|x|}}$ is
- (a) continuous everywhere but not differentiable at $x = 0$ ✓ Correct
- (b) continuous and differentiable everywhere
- (c) not continuous at $x = 0$
- (d) None of the above
The function $f(x) = {e^{|x|}}$ is
Let $u(x) = |x|$ and $v(x) = {e^x}$
therefore,$f(x) = {\mathop{\rm vou}\nolimits} (x) = v[u(x)]$
$= v|x| = {e^{|x|}}$
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 $x = 0$.
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