Question

The function f (x) = e^−|x| is

Options

• continuous everywhere but not differentiable at x = 0

• continuous and differentiable everywhere

• not continuous at x = 0

• none of these

Answer 1

continuous everywhere but not differentiable at x = 0 

Given:

`f(x) = e^(-|x|) = {(e^x , xge0),(e^(-x) ,x< 0):}`
\[\text{ Continuity }: \]
\[ \lim_{x \to 0^-} f(x) \]
\[ = \lim_{h \to 0} f(0 - h) \]
\[ = \lim_{h \to 0} e^{- (0 - h)} \]
\[ = \lim_{h \to 0} e^h \]
\[ = 1\]

RHL at x = 0

\[\lim_{x \to 0^+} f(x) \]
\[ = \lim_{h \to 0} f(0 + h) \]
\[ = \lim_{h \to 0} e^{(0 + h)} \]
\[ = 1\]

and f(0) =

\[f(0) = e^0 = 1\]

Thus,

\[\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f\]

Hence, function is continuous at x = 0
Differentiability at x = 0

(LHD at x = 0)

\[\lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0}\]
\[ = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{0 - h - 0}\]
\[ = \lim_{h \to 0} \frac{e^{- (0 - h)} - 1}{- h}\]
\[ = \lim_{h \to 0} \frac{e^h}{h} \]
\[ = \infty\]

Therefore, left hand derivative does not exist.
Hence, the function is not differentiable at x = 0.

Answer 2

continuous everywhere but not differentiable at x = 0

Explanation:

1. Continuity: The function f(x) = e^−∣x∣ is continuous everywhere because both e^−x and e^x are continuous functions, and the absolute value function ∣x∣ does not introduce any discontinuities.
2. Differentiability: At x = 0, f(x) is not differentiable because the derivative from the left and the right do not match.
   • For x > 0: f′(x) = −e^−x
   • For x < 0: f′(x) = e^x
   • At x = 0: The derivative changes abruptly due to the absolute value term, making the function non-differentiable at x = 0.

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