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Question

Examine the following curve for continuity and differentiability: y=x2 for x0;y=1 for0x1 andy=1/x for x>1. Also draw the graph of the function.

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Solution

f(x)=⎪ ⎪⎪ ⎪x2x010<x11xx>1

f(x)=x2 for x<0 is continuous, being a polynomial function.
f(x)=1 for x(0,1) is continuous, being a constant function.
f(x)=1x for x>1 , again a part of reciprocal function, so continuous.

We need to check continuity at x=0 and x=1
At x=0
LHL=limx0f(x)
=limh0f(0h)
=limh0(h)2
LHL=0

RHL=limx0+f(x)
=limh0f(0+h)
RHL=1

f(0)=1
Since, LHLRHL.
So, the function is discontinuous at x=0

At x=1
LHL=limx1f(x)
=limh0f(1h)
LHL=1

RHL=limx1+f(x)
=limh0f(1+h)
=limh011+h
RHL=1

f(1)=1
Since, LHL=RHL=f().
So, the function is continuous at x=1

Since, f(x) is not continuous at x=0
So, f(x) is also not differentiable at x=0

Now, we will check differentiability at x=1
Lf(1)=limh0f(1h)f(1)h
=limh011h
Lf(1)=0

Rf(1)=limh0f(1+h)f(1)h
=limh011+h1h
Rf(1)=0

Since, Lf(1)=Rf(1)
Hence, f(x) is differentiable at x=1

409921_165339_ans.jpg

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