First slide

Derivatives

Question

$If f (x) = \{\begin{matrix} \frac{x (3 e^{1 / x} + 4)}{2 - e^{1 / x}}, x \neq 0 \\ 0, x = 0 \end{matrix}, then f (x) is$

Moderate

A

Continuous as well as differentiable at x=0
B

Continuous but not differentiable at x=0
C

Differentiable but not continuous at x=0
D

Discontinuous every where

Solution

$\begin{array}{l} L f^{1} (0) = l t \frac{f (0 - h) - f (0)}{- h} \\ = l t_{n \to 0} (\frac{- h (3 e^{- 1 / h} + 4)}{2 - e^{- 1 / h}} - 0) (\frac{- 1}{h}) = 2 \\ R f^{1} (0) = {lt}_{h \to 0} \frac{f (0 + h) - f (0)}{h} \\ = l t_{h \to 0} (\frac{h (3 e^{1 / h} + 4)}{2 - e^{1 / h}} - 0) (\frac{1}{h}) \\ = {lt}_{h \to 0} (\frac{3 + 4 e^{- 1 / h}}{2 e^{- 1 / h} - 1}) = - 3 \\ Since L f^{1} (0) \neq R f^{1} (0) \end{array}$
$∴ f (x) is not differentiable at x = 0 . But f (x) is continuous at x = 0$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App