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Question: The set of all points where the function f(x) = \(\sqrt{1 - e^{- x^{2}}}\)is differentiable is...

The set of all points where the function f(x) = 1−e−x2\sqrt{1 - e^{- x^{2}}}is

differentiable is

A

(0 , ¥)

B

( –¥, ¥)

C

(–¥, ¥) ~ {0}

D

(–1, ¥)

Answer

(–¥, ¥) ~ {0}

Explanation

Solution

For x ¹ 0, we have

f¢(x) = 12\frac { 1 } { 2 } 11−e−x2\frac{1}{\sqrt{1 - e^{- x^{2}}}} [–(–2x) e−x2e^{- x^{2}}]

= xe−x21−e−x2\frac{xe^{- x^{2}}}{\sqrt{1 - e^{- x^{2}}}}

Also, f¢(0+) = f(h)−f(0)h\frac{f(h) - f(0)}{h}

= 1−e−h2h\frac{\sqrt{1 - e^{- h^{2}}}}{h}

= (e−h2−1−h2)1/2\left( \frac{e^{- h^{2}} - 1}{- h^{2}} \right)^{1/2}= 1

and f¢(0–) =lim⁡h→0–\lim_{h \rightarrow 0^{–}} – (e−h2−1−h2)1/2\left( \frac{e^{- h^{2}} - 1}{- h^{2}} \right)^{1/2}= –1

because, as h ® 0– , h is a negative number, so that

1−e−h2h\frac{\sqrt{1 - e^{- h^{2}}}}{h}= – 1−e−h2∣h∣\frac{\sqrt{1 - e^{- h^{2}}}}{|h|}= – 1−e−h2h2\sqrt{\frac{1 - e^{- h^{2}}}{h^{2}}}

= – (e−h2−1−h2)1/2\left( \frac{e^{- h^{2}} - 1}{- h^{2}} \right)^{1/2}

Hence f is not differentiable at x = 0 .

Thus the points of differentiability are (–¥, ¥) ~ {0}