Q.The function $f(x) = e^{|x|}$ is
(A) continuous everywhere but not differentiable at $x = 0$
(B) continuous and differentiable everywhere
(C) not continuous at $x = 0$
(D) none of these

The key idea is that $|x|$ has a sharp corner at $x=0$, making it non-differentiable there, but the exponential function is smooth and preserves continuity.

1. Continuity: $|x|$ is continuous everywhere, and $e^t$ is continuous for all real $t$. The composition $e^{|x|}$ is therefore continuous everywhere, including at $x=0$.

2. Differentiability at $x=0$: Check the left and right derivatives.

   For $x>0$, $f(x)=e^x$, so $f'(x)=e^x$ and $f'(0^+)=1$.

   For $x<0$, $f(x)=e^{-x}$, so $f'(x)=-e^{-x}$ and $f'(0^-)=-1$. …

The key to this problem is understanding how the absolute value function behaves inside another function. $|x|$ is continuous everywhere but has a corner at $x=0$ — its derivative jumps from $-1$ to $+1$. When you wrap that inside $e^u$, which is smooth and strictly increasing, the corner is preserved: the exponential stretches the values but does not smooth out the kink.

Let’s walk through the reasoning step by step.

1. Check continuity at $x=0$.

   The function $f(x) = e^{|x|}$ is a composition of $|x|$ (continuous everywhere) and $e^u$ (continuous everywhere). A composition of continuous functions is continuous.

   At $x=0$, we have $f(0) = e^{0} = 1$. The left-hand limit: $\lim_{x \to 0^-} e^{|x|} = e^{0} = 1$. The right-hand limit: $\lim_{x \to 0^+} e^{|x|} = e^{0} = 1$. All three match, so $f$ is continuous at $0$.

   Note

   Continuity is never the issue here — the absolute value function is continuous, and exponentiating preserves that.

2. Now test differentiability at $x=0$.

   Differentiability requires that the left-hand derivative equals the right-hand derivative. For $x < 0$, $|x| = -x$, so $f(x) = e^{-x}$. For $x > 0$, $|x| = x$, so $f(x) = e^{x}$.

   Compute the left-hand derivative at $0$:

Using the standard limit $\lim_{t \to 0} \frac{e^{t} - 1}{t} = 1$, with $t = -h$, we get:

Compute the right-hand derivative at $0$:

Since $f'_-(0) = -1 \neq 1 = f'_+(0)$, the derivative does not exist at $x=0$. …

Method: Continuity and Differentiability of $g(|x|)$ at $x=0$

Use this whenever a smooth function $g$ is applied to $|x|$, creating a possible corner at the origin.

Steps

Step 1: Establish continuity first, separately from differentiability.

$|x|$ is continuous everywhere, and if $g$ is continuous everywhere (like $g(u)=e^u$), the composition $g(|x|)$ is continuous everywhere too — composition of continuous functions is continuous. This step usually settles quickly and rules out any 'discontinuous' option.

Step 2: Split into two smooth pieces using $|x| = x$ for $x>0$ and $|x|=-x$ for $x<0$.

Step 3: Differentiate each piece using the chain rule, then take the limit of each formula as $x \to 0$ from its own side. …

Common Mistakes

Mistake 1: Differentiating $e^{|x|}$ using the single formula $e^{|x|}\cdot\frac{x}{|x|}$ and then substituting $x=0$ directly.

Why it's wrong: $\frac{x}{|x|}$ itself is undefined at $x=0$ (it jumps between $-1$ and $+1$), so plugging in $x=0$ gives a meaningless or misleading result instead of revealing the actual one-sided behaviour.

Correct approach: split into the two smooth pieces $e^{x}$ (for $x>0$) and $e^{-x}$ (for $x<0$), and evaluate the left-hand and right-hand derivatives separately using the limit definition at $x=0$.

Mistake 2: Assuming that because $e^{|x|}$ is continuous everywhere, it must also be differentiable everywhere. …