4.
Lefthand derivative:
lim
h
→
0
2
}
f
(1
1
h
h
)
2
f
(1)
}5
lim
h
→
0
2
}
(1
1
h
h
)
2
1
}5
lim
h
→
0
2
1
5
1
Righthand derivative:
lim
h
→0
1
}
f
(1
1
h
h
)
2
f
(1)
}5
lim
h
→0
1
5
lim
h
→0
1
}
1
h
2
(1
(1
1
1
h
)
h
)
}
5
lim
h
→0
1
}
h
(1
2
1
h
h
)
}
5
lim
h
→0
1
2}
1
1
1
h
}52
1
Since 1
±2
1, the function is not differentiable at the
point
P
.
5. (a)
All points in [
2
3, 2]
(b)
None
(c)
None
6. (a)
All points in [
2
2, 3]
(b)
None
(c)
None
7. (a)
All points in [
2
3, 3] except
x
5
0
(b)
None
(c)
x
5
0
8. (a)
All points in [
2
2, 3] except
x
52
1, 0, 2
(b)
x
52
1
(c)
x
5
0,
x
5
2
9. (a)
All points in [
2
1, 2] except
x
5
0
(b)
x
5
0
(c)
None
10. (a)
All points in [
2
3, 3] except
x
52
2, 2
(b)
x
52
2,
x
5
2
(c)
None
11.
Since lim
x
→
0
tan
2
1
x
5
tan
2
1
0
5
0
±
y
(0), the problem is a
discontinuity.
12.
lim
h
→
0
2
}
y
(0
1
h
h
)
2
y
(0)
}5
lim
h
→
0
2
}
h
h
4/5
} 5
lim
h
→
0
2
}
h
1
1/5
}52‘
lim
h
→
0
1
}
y
(0
1
h
h
)
2
y
(0)
}5
lim
h
→
0
1
}
h
h
4/5
} 5
lim
h
→
0
1
}
h
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.
 Spring '08
 GERMAN
 Derivative

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