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  • Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.

 Q21  Does there exist a function which is continuous everywhere but not differentiable
         at exactly two points? Justify your answer.

Answers (1)

best_answer

Consider f(x) = |x| + |x +1|
We know that modulus functions are continuous everywhere and sum of two continuous function is also a continuous function
Therefore, our function f(x) is continuous
Now,
If Lets differentiability of our function at x = 0 and x= -1
L.H.D. at x = 0
\lim_{h\rightarrow 0^-}\frac{f(x+h)-f(x)}{h}= \lim_{h\rightarrow 0^-}\frac{f(h)-f(0)}{h}= \lim_{h\rightarrow 0^-}\frac{|h|+|h+1|-|1|}{h}
                                                                                            =\lim_{h\rightarrow 0^-}\frac{-h-(h+1)-1}{h}= 0                      (|h| = - h \ because\ h < 0)
R.H.L. at x = 0
\lim_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h}= \lim_{h\rightarrow 0^+}\frac{f(h)-f(0)}{h}= \lim_{h\rightarrow 0^+}\frac{|h|+|h+1|-|1|}{h}
                                                                                            =\lim_{h\rightarrow 0^+}\frac{h+h+1-1}{h}= \lim_{h\rightarrow 0^+}\frac{2h}{h}= 2          (|h| = h \ because \ h > 0)
R.H.L. is not equal to L.H.L.
Hence.at  x = 0 is the function is not differentiable
Now, Similarly
R.H.L. at x = -1
\lim_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h}= \lim_{h\rightarrow 0^+}\frac{f(-1+h)-f(-1)}{h}= \lim_{h\rightarrow 0^+}\frac{|-1+h|+|h|-|-1|}{h}
                                                                                                            =\lim_{h\rightarrow 0^+}\frac{1-h+h-1}{h}= \lim_{h\rightarrow 0^+}\frac{0}{h}= 0                                                                                                                                                                                                                                                                            (|h| = h \ because \ h > 0)
L.H.L. at x = -1
\lim_{h\rightarrow 0^-}\frac{f(x+h)-f(x)}{h}= \lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h}= \lim_{h\rightarrow 0^-}\frac{|-1+h|+|h|-|1|}{h}
                                                                                                     =\lim_{h\rightarrow 1^+}\frac{1-h-h-1}{h}= \lim_{h\rightarrow 0^+}\frac{-2h}{h}= -2       (|h| = - h \ because\ h < 0)
L.H.L. is not equal to R.H.L, so not differentiable at x=-1

Hence, exactly two points where it is not differentiable
 

Posted by

Gautam harsolia

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