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Q.Can we say that all even functions are continuous and non differentiable at x=0 and all odd functions are continuous and differentiable at x=0:?

siddharth gupta , 10 Years ago
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anser 3 Answers
Arun Kumar

Last Activity: 10 Years ago

Hello Student,

No we can’t generally say that.
In even function take the example of cosx
its continous and differentiable at 0.
In odd function take example of sin(1/x).
its non continous and non differentiable at 0.

Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty

Jitender Singh

Last Activity: 10 Years ago

Ans: Yes
Sol:
Let f(x) be the even function, We have
f(x)= f(-x)
f(0)= 0
Proof for continuity:
Right Hand Limit (RHL) = f(0) = Left Hand Limit (LHL)
\lim_{h\rightarrow 0}f(h) = 0 = \lim_{h\rightarrow 0}f(-h)
RHL = \lim_{h\rightarrow 0}f(h) = f(0) = 0
LHL = \lim_{h\rightarrow 0}f(-h) = f(0) = 0
Proof for Non Differentiability:
For f to be differentiable,
Left Hand Derivative (LHD) = Right Hand Derivative (RHD)
\lim_{h\rightarrow 0}\frac{f(-h)-f(0)}{-h} = \lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h}
LHD = \lim_{h\rightarrow 0}\frac{f(-h)-f(0)}{-h} = \lim_{h\rightarrow 0}-\frac{f(h)}{h}
RHD = \lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h} = \lim_{h\rightarrow 0}\frac{f(h)}{h}
LHD \neq RHD
Let f(x) be odd,
f(x)= -f(-x)
f(0)= 0
Proof for continuity:
RHL = f(0) = LHL
RHL = \lim_{h\rightarrow 0}f(h) = f(0) = 0
LHL = \lim_{h\rightarrow 0}f(-h) = f(0) = 0
Proof for Differentiability:
LHD = RHD
LHD = \lim_{h\rightarrow 0}\frac{f(-h)-f(0)}{-h} = \frac{-f(h)}{-h} = \frac{f(h)}{h}
RHD = \lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h} = \frac{f(h)}h}
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

Jitender Singh

Last Activity: 10 Years ago

There is a little mistake in the previous solution.
In case of even, differentiability depend on the f(x). For example,
f(x) = x2,  LHD = RHD = 0
f(x) = |x|, LHD = -1   RHD = 1
x2 is differentiable but |x| is not.
Similarly for the odd function. For example,
f(x) = x3, LHD = 0 RHD = 0  Differentiable
f(x) = x1/3, Derivative becomes infinite. So non differentiable
 
 
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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