First slide

Differentiability

Question

$If f (x) = \{\begin{matrix} x \frac{e^{(1 / x)} - e^{(- 1 / x)}}{e^{(1 / x)} + e^{(- 1 / x)}} & ; x \neq 0 \\ 0 & ; x = 0 \end{matrix}, then which of the following is true$

Difficult

A

f is continuous and differentiable at every point
B

f is continuous at every point but is not differentiable everywhere
C

f is differentiable at every point
D

f is differentiable only at the origin

Solution

$f (0 -) = \lim_{x \to 0^{-}} x \frac{e^{\frac{1}{x}} - e^{- \frac{1}{x}}}{e^{\frac{1}{x}} + e^{- \frac{1}{x}}}$
$= \lim_{x \to 0^{-}} x \frac{e^{\frac{2}{x}} - 1}{e^{\frac{2}{x}} + 1} = \lim_{x \to 0} x (\frac{0 - 1}{0 + 1}) = 0$
$f (0 +) = \lim_{x \to 0^{+}} x \frac{1 - e^{- \frac{2}{x}}}{1 + e^{- \frac{2}{x}}} = \lim_{x \to 0^{+}} x (\frac{1 - 0}{1 + 0}) = 0$
$f (0) = 0 (given)$
So, f is continuous everywhere
$f^{'} (0 -) = \lim_{k \to 0^{-}} \frac{h (\frac{e^{\frac{1}{h}} - e^{- \frac{1}{h}}}{e^{\frac{1}{h}} + e^{- \frac{1}{h}}}) - 0}{h} = \frac{0 - 1}{0 + 1} = - 1$
$f' (0 +) = \lim_{h \to 0^{+}} \frac{h (\frac{e^{\frac{1}{h}} - e^{- \frac{1}{h}}}{e^{\frac{1}{h}} + e^{- \frac{1}{h}}}) - 0}{h} =1 ∴ f' (0 -) \neq f' (0 +)$

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