Limits Continuity And Differentiability Question 66
Question: A function $ f:R\to R $ is defined as $ f(x)=x^{2} $ for $ x\ge 0 $ and $ f(x)=-x $ for $ x<0 $ . Consider the following statements in respect of the above function:
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The function is continuous at x = 0.
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The function is differentiable at x = 0.
Which of the above statements is/are correct?
Options:
A) 1 only
B) 2 only
C) Both 1 and 2
D) Neither 1 nor 2
Show Answer
Answer:
Correct Answer: A
Solution:
$ f:R\to R,f(x)= \begin{matrix} x^{2}, & x\ge 0 \\ -x, & x<0 \\ \end{matrix} . $ For continuity at x = 0 $ f(0-0)=\underset{h\to 0}{\mathop{\lim }}f(0-h) $
$ =\underset{h\to 0}{\mathop{\lim }}[(0-h)]=\underset{h\to 0}{\mathop{\lim }}h=0 $
$ f(0+0)=\underset{h\to 0}{\mathop{\lim }}f(0+h)=\underset{h\to 0}{\mathop{\lim }}{{(0+h)}^{2}}=0 $ and $ f(0)=0 $
$ f(x) $ is continuous at x = 0
For differentiability at $ x=0 $
$ \underset{h\to 0}{\mathop{\lim }}\frac{-(-h)-0}{-h}=\underset{h\to 0}{\mathop{\lim }}=\frac{h}{-h}=-1 $ and $ \underset{h\to 0}{\mathop{\lim }}\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}h=0 $
$ f(x) $ is not differentiable at x ? 0
BETA
