class="bi x0 y0 w1 h1"
y =
ax
2
+ bx + c
dx + e
y =
x
2
− mx + 1
x − 1
y =
x
2
− mx + 1
x − 1
(∗)
y = ax
3
+ bx
2
+ cx + d
class="bi x1 y1 w3 h6"
y = f(x) x → x
0
y = f(x) x
0
x
0
∀ε > 0, ∃δ = δ(ε) : 0 < |x − x
0
); f(x
2
); f(x
3
) ; f(x
n
)
lim
x→0
x.sin
1
x
= 0
f(x) = x.sin
1
x
x
0
= 0
x
0
= 0 x
n
(
−π
4
;
π
4
)
x→0
x.sin
1
x
= 0
lim
x→1
sin
1
x − 1
x
n
= 1 +
1
nπ
n
= 1 +
2
(4n + 1)π
lim
n→∞
x
n
= 1; lim
n→∞
n
= 1
f(x
n
) = sin
x − 1
y = f(x
n
) x
0
x → x
0
A =
lim
x→(x
0
+0)
f(x) = f(x
0
+ 0) ∀ε > 0, ∃δ = δ(ε) > 0 ∀x : 0 < x − x
0
< δ ⇒
|f(x) − A| < ε
y = f(x) x
0
x
0
f(x) x → x
0
A = lim
x→(x
0
−0)
f(x) = f(x
0
) ∃ M > 0
|f(x)| ≤ M, ∀x ∈ V (x
0
), x = x
0
V (x
0
)
1 > |f(x) − A| ≥ |f(x)| − |A|
⇒ |f(x)| < 1 + |A| 1 +|A|
lim
x→x
0
f(x) = A, A = 0
V (x
0
) |f(x)| >
|A|
2
, ∀x ∈ V (x
0
), x = x
0
lim
x→x
0
f
1
(x) = A
1
2
(x) = A f
1
(x) ≤ ϕ(x) ≤
f
2
(x), ∀x ∈ V (x
0
), x = x
0
lim
x→x
0
ϕ(x) = A
∃ lim
x→x
0
f(x)
y = f(x) x
0
x
0
∀ε > 0
∃ V (x
0
) |f(x
) − f(x”)| < ε ∀x
, x” ∈ V (x
x
= 1
f(x) =
sinx
x
x
0
= 0
V (x
0
) = x : 0 < |x| <
π
2
0 < x <
π
2
S
AOM
< S
quatAOM
< S
AOT
⇔
1
2
OA.MH <
1
2
.OA
2
< x < 0
lim
x→0
1
cosx
= 1 ⇒ lim
x→0
sinx
x
= 1
x
0
x
0
lim
x→x
0
= f(x
0
) x
0
x
0
x
0
x
0
f(x
0
) = 0 c = u
0
f(u
0
< 0)
α
1
= u
0
, β
1
= β
0
f(u
0
> 0) α
1
= α
0
, β
1
= u
0
[α
1
, β
1
] f(α
1
).f(β
, β
n+1
= β
n
f(u
n
> 0)
α
n+1
= α
n
, β
n+1
= u
n
α
n
, β
n
f(x)
µ ∈ [m, M]
m = min f(x), M = maxf (x) ∃ξ ∈ (a, b) : f (ξ) = µ
y = f(x) f(x)
c ∈ (a, b)
f
(c) f
(c) = 0
y = f(x) [a; b]
∈ [2; 3]; ; c
2013
∈ [2013; 2014]
f
(c
1
) = 0, f
(c
2
) = 0, , f
(c
2013
) = 0
⇒ f
(x) ⇒
f
(x) = 0 C
1
; C
2
; C
3
; ; C
2013
f(x) = x
∃c ∈ (a, b) h
(c) = 0
h
(x) = f
(x) −
f(b) − f(a)
b − a
⇒ h
(c) = f
(c) −
f(b) − f(a)
b − a
= 0 →
f
(c) =
f(b) − f(a)
b − a
, c ∈ (a, b)
0 < b < a
a − b
a
< ln
a
b
<
ln
a
b
a − b
<
1
b
⇒
a − b
a
< ln
a
b
<
a − b
b
y = f(x)
[a, f(a)] , [b, f(b)]
y = f(x), y = g(x)
g
(x) = 0, ∀x ∈ (a, b)
∃c ∈ (a, b) :
f(b) − f(a)
b − a
=
f
(c)
⇒
f
(c)
g
(c)
=
f(b) − f(a)
g(b) −g(a)
y =
ax
2
+ bx + c
dx + e
y =
x
2
− mx + 1
x − 1
m = 1
y =
x
2
− x + 1
x − 1
= x +
1
x − 1
. (∗∗)
y = 0 → x
2
−x + 1 = 0 →
x = 0 → y = −1 →
→
x = X + 1
y = Y + 1
Y + 1 = X + 1 +
1
X + 1 −1
⇔ Y = X +
1
X
1
x − 1
⇒ x − 1 = ±1 ⇒
x
1
= 1, y
1
= −1
x
2
= 2, y
2
= 3.
M
1
− x + 1
x − 1
= −2x ⇔ 3x
2
−3x + 1 = 0
⇒
x
2
− x + 1
x − 1
= 2x ⇔ x
2
− x − 1 = 0
⇒ x =
1 ±
√
5
2
x =
1 ±
√
5
2
⇒
d
1
= OD =
(x
2
0
+ (−1)y
0
|
1
2
+ (−1)
2
=
1
√
2
.|x
0
− (x
0
+
1
x
0
− 1
)| =
1
√
2
.
1
|x
0
0
− 1|
=
2
4
√
2
|x
0
− 1| =
1
√
2.|x
0
− 1|
⇔ (x
0
− 1)
2
=
1
√
2
⇒ x
0
= 1 ±
1
4
√
2
1
β
MN =
(x
M
− x
N
)
2
+ (y
M
− y
N
)
2
(α + β)
2
+ (α + β +
1
α
+
1
β
)
2
(α + β).
+ 2 2
(2.
√
2 + 2)
⇔ α = β =
1
4
√
2
x
M
= 1 +
1
4
√
2
x
N
= 1 −
1
4
√
2
M(x
0
, y
0
)
d
√
2
.
1
|x
0
− 1|
d
1
.d
2
=
1
√
2
M(x
0
, y
0
)
⇒
(x
0
− 1)
2
+ (y
0
− 1)
2
2
(2(x
0
− 1)
2
.
1
(x
0
− 1)
2
+ 2
⇔ 2(x
0
− 1)
2
=
1
(x
0
− 1)
2
⇔ (x
0
− 1)
4
=
1
2
⇒ Y = X +
1
X
. (i)
⇒
⇒ M
,
(−X, −Y )
M
,
M
,
⇔
M
,
⇔ Y (−X) = Y
,
(X)
Y
,
(X) = 1 −
1
X
2
= 1 −
1
(−X)
2
= Y
,
0
)
⇔ y =
1 −
1
(x
0
− 1)
2
.x +
x
0
(x
0
− 1)
2
+
1
x
0
− 1
⇔ y =
1 −
1
(x
0
− 1)
E
1, 1 +
2
X
0
− 1
⇒ IE =
2
|x
0
− 1|
F (2x
0
− 1, 2x
0
− 1)
⇒ IF =
(2x
0
− 2)
2
+ (2x
0
− 2)
2
=
0
⇒ y
,
(x
0
) = −2 ⇔ 1 −
1
(x
0
− 1)
2
= −2 ⇔ (x
0
− 1)
2
=
1
3
⇒ x
0
= 1 ±
1
√
3
x
0
= 1 ±
1
√
3
0
, 0) ∈
(x
0
, 0) y = k(x − x
0
)
x
2
− x − 1
x − 1
= kx − kx
0
⇔ f(x) = (k − 1)x
2
− (k − kx
0
+ 1)x + kx
0
− 1 = 0
x = 1 ⇒ = 0
⇔ (k + kx
0
− 1)
2
− 4(k −1)(kx
0
− 1) = 0
⇔ (x
0
= −1 ⇒
−3
(x
0
− 1)
2
= −1 ⇔ (x
0
− 1)
2
= 3 ⇒ x
0
= 1 ±
√
3
x
0
= 1 ±
√
3
(x
0
, 0) ∈
(x
0
, 0) y = k(x − x
0
)
x
2
2
+ 4k + 4kx
0
− 4 = 0
⇔ (x
0
− 1)
2
k
2
+ 2(x
0
+ 1)k −3 = 0 (i)
x
0
= 1 ⇒ ∃ k =
3
4
x
0
= 1
,
≥ 0
⇔ (x
0
+ 1)
2
+ 3(x
0
− 1)
0
−
1
x
0
− 1
= 1 − x
0
+
x
0
− 1
(x
0
− 1)
2
⇔ −
1
x
0
− 1
=
1
x
0
− 1
⇔
2
x
0
⇒
k = 1
(k −1)f (−1) > 0
1 <
S
2
⇔
k = 1
(k −1)(−1) > 0
1 <
−2
2(k −1)
k = 1
(k −1)f (1) > 0
S
2
< 1
⇔
k = 1
(k −1)(−1) > 0
−2
2(k −1)
< 1
⇔
k = 1
∀k
1 + (k −1)(k + 2) = 0
⇔
k = 1
k
2
+ k −1 = 0
⇔ k =
−1 ±
√
5
2
f(x, y) = kx −y + k + 1 = 0
⇔ −2 < k <
2
3
⇔ k < −2
k >
2
3
⇔ k = −2
k =
x − 1
− log
2
a = 0
x
2
− x + 1
x − 1
= log
2
(a)
y = log
2
a
≤ 0 ⇒ ∃log
2
a ⇒
log
2
(a) < −1 = log
2
(2)
−1
= log
2
(
1
2
⇔ 0 < a <
1
log
2
a = 3 ⇔ a = 8 ⇒ y = log
2
a
⇒
log
2
a > 3 ⇔ a > 8 ⇒
⇒
x
2
− x + 1
x − 1
=
a
2
− a + 1
a − 1
−∞ < a < 1
a = 0
⇒
a
2
− a + 1
a − 1
< −1
⇒ y =
a
2
− a + 1
a − 1
⇒
45
0
⇒ 45
0
y
,
(x) = ±1 ⇒ 1 −
1
(x − 1)
2
= ±1 ⇒ 1 −
1
(x − 1)
2
= −1
⇔ 2 =
1
(x − 1)
2
⇔ (x − 1)
2
=
1
2
⇒ x = 1 ±
1
√
2
)
x
2
− x + 1
x − 1
= 6 x
A
, x
B
: |x
A
− x
B
| = 6
⇔ x
2
−(b+1)x+1+b = 0 x
A
, x
B
: |x
A
−x
B
| = 6
⇔ x
A
, x
y =
x
2
− 2x + 2
x − 1
. (33)
⇔ y =
x
2
− x + 1
x − 1
− 1.
y =
x
2
x − 1
. (34)
⇔ y =
x
2
− x + 1
x − 1
+ 1
y =
x
2
− 3x + 3
x − 2
. (35)
⇔ y =
x
2
− x + 1
x − 1
,
x
2
− x + 1
x − 1
> 0
−
x
2
− x + 1
x − 1
,
x
2
− x + 1
x − 1
< 0
⇔ y =
x
2
− x + 1
x − 1
, x ≥ 0
x
2
+ x + 1
−x − 1
, x < 0
|y| =
x
2
− x + 1
x − 1
. (39)
x
2
− x + 1 > 0, ∀x ⇒ (39) ⇔ x − 1 > 0 ⇔ x > 1
⇔ y = ±
x
2
− x + 1
x − 1
,
x
2
− x + 1
x − 1