7.Průběh funkce-příklady
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2
(1 + x2)2
= 0
1 − x
2
= 0
x2 = 1
x1 = 1
x2 = −1
&
−1
min
%
1
MAX
&
y00 =
1 − x
2
(1 + x2)2
!
0
=
−2x(1 + x
2)2−(1 − x2)2(1 + x2)(0 + 2x)
(1 + x2)4
=
−2x(1 + x
2)[(1 + x2) + (1 − x2)2]
(1 + x2)4
=
−2x[3 − x
2]
(1 + x2)3
= 2
x
(x
2 − 3)
(1 + x2)3
y00 = 2
x
(x
2 − 3)
(1 + x2)3
⇒
2
x
(x
2 − 3)
(1 + x2)3
= 0
⇒
x
(x2 − 3) = 0
x3 = 0, x4 =
p
3, x5 = −
p
3
∩
−
√
3
in.
∪
0
in.
∩
√
3
in.
∪
Uprav´ıme.
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y =
x
1 + x2
D
(f ) = R; lich ´a;
−
0
+
y0 =
1 − x
2
(1 + x2)2
;
x1,2 = ±1
y0 = 0
1 − x
2
(1 + x2)2
= 0
1 − x
2
= 0
x2 = 1
x1 = 1
x2 = −1
&
−1
min
%
1
MAX
&
y00 =
1 − x
2
(1 + x2)2
!
0
=
−2x(1 + x
2)2−(1 − x2)2(1 + x2)(0 + 2x)
(1 + x2)4
=
−2x(1 + x
2)[(1 + x2) + (1 − x2)2]
(1 + x2)4
=
−2x[3 − x
2]
(1 + x2)3
= 2
x
(x
2 − 3)
(1 + x2)3
y00 = 2
x
(x
2 − 3)
(1 + x2)3
⇒
2
x
(x
2 − 3)
(1 + x2)3
= 0
⇒
x
(x2 − 3) = 0
x3 = 0, x4 =
p
3, x5 = −
p
3
∩
−
√
3
in.
∪
0
in.
∩
√
3
in.
∪
Hled ´ame ˇreˇsen´ı rovnice y0 = 0.
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y =
x
1 + x2
D
(f ) = R; lich ´a;
−
0
+
y0 =
1 − x
2
(1 + x2)2
;
x1,2 = ±1
y0 = 0
1 − x
2
(1 + x2)2
= 0
1 − x
2
= 0
x2 = 1
x1 = 1
x2 = −1
&
−1
min
%
1
MAX
&
y00 =
1 − x
2
(1 + x2)2
!
0
=
−2x(1 + x
2)2−(1 − x2)2(1 + x2)(0 + 2x)
(1 + x2)4
=
−2x(1 + x
2)[(1 + x2) + (1 − x2)2]
(1 + x2)4
=
−2x[3 − x
2]
(1 + x2)3
= 2
x
(x
2 − 3)
(1 + x2)3
y00 = 2
x
(x
2 − 3)
(1 + x2)3
⇒
2
x
(x
2 − 3)
(1 + x2)3
= 0
⇒
x
(x2 − 3) = 0
x3 = 0, x4 =
p
3, x5 = −
p
3
∩
−
√
3
in.
∪
0
in.
∩
√
3
in.
∪
Dosad´ıme za derivaci.
//
/
.
..
c
Robert Maˇr´ık, 2008 ×