() = {− 1 − 2 < <− 1 1 + − 1≤≤0 1 −
2
0≤ < 1 1 1≤ < 2 2 = 2 3 − 2
Sketch the graph of f(x). Discuss about existence of left limit, right limit, limit of f(x) at
x = -1,0,1,2.
Answers
Answer:
1
Explanation:
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Class 12
>>Maths
>>Continuity and Differentiability
>>Differentiability of a Function
>>Examine the following curve...
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Examine the following curve for continuity and differentiability: y=x
2
for x≤0;y=1 for0x≤1 andy=1/x for x>1. Also draw the graph of the function.
Hard
Solution
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f(x)=
⎩
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎧
x
2
x≤0
10<x≤1
x
1
x>1
f(x)=x
2
for x<0 is continuous, being a polynomial function.
f(x)=1 for x∈(0,1) is continuous, being a constant function.
f(x)=
x
1
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=lim
x→0
−
f(x)
=lim
h→0
f(0−h)
=lim
h→0
(−h)
2
⇒LHL=0
RHL=lim
x→0
+
f(x)
=lim
h→0
f(0+h)
⇒RHL=1
f(0)=1
Since, LHL
=RHL.
So, the function is discontinuous at x=0
At x=1
LHL=lim
x→1
−
f(x)
=lim
h→0
f(1−h)
⇒LHL=1
RHL=lim
x→1
+
f(x)
=lim
h→0
f(1+h)
=lim
h→0
1+h
1
⇒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)=lim
h→0
−h
f(1−h)−f(1)
=lim
h→0
−h
1−1
⇒Lf
′
(1)=0
Rf
′
(1)=lim
h→0
h
f(1+h)−f(1)
=lim
h→0
h
1+h
1
−1
⇒Rf
′
(1)=0
Since, Lf
′
(1)=Rf
′
(1)
Hence, f(x) is differentiable at x=1