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Math 101 Chapter 4: Taylor series and complex numbers
4.20 Systematic curve sketching
4.22 On using the sign tests
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4.21 Stationary points
The derivatives are
f
(
x
)
=
x
+
(
x
+
1
)
-
1
f(x)=x+(x+1)^{-1}
f
′
(
x
)
=
1
-
(
x
+
1
)
-
2
f^{\prime}(x)=1-(x+1)^{-2}
f
′′
(
x
)
=
2
(
x
+
1
)
-
3
;
f^{\prime\prime}(x)=2(x+1)^{-3};
so stationary points occur where
f
′
(
x
)
=
0
;
f^{\prime}(x)=0;
that is
(
x
+
1
)
2
=
1
,
so
x
+
1
=
±
1.
(x+1)^{2}=1,\quad{\hbox{so}}\quad x+1=\pm 1.
The nature of the stationary points
x
f
(
x
)
f
′′
(
x
)
nature of point
x\qquad f(x)\qquad f^{\prime\prime}(x)\qquad\qquad{\hbox{nature of point}}
0
1
2
local minimum
0\qquad 1\quad\qquad 2\quad\qquad\qquad{\hbox{local minimum}}
-
2
-
3
-
2
local maximum
-2\quad-3\quad\quad-2\quad\qquad\qquad{\hbox{local maximum}}