Επιμέλεια: ΕΜΕ Ημαθίας αποκλειστικά για το lisari.blogspot.com

1. - µ
2019
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 1 of 17

2. 1
f : R ! R , µ
:
f 0
(x)° f (x)
f (x)
=
x°1
ex ° x
, 8x 2 R (1)
f (0) = 1 ∑Æ∂ f (x) ∏ 1, 8x 2 R (2)
1. f (x).
2. : f (x)+ f (°x) ∏ 2. " = "
3. µ µ
µ .
4. : ex
= x+ a
µ a 2 R.
5. µ , µ µ
f (x).
6. µ (")
f (x) .
7. µ µ -
f , µ (")
x = 0.
8. ex
∏ ex, x 2 R.
1
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 2 of 17

3. 2
1. 1 µ :
f 0
(x)°f (x)
f (x) = x°1
ex°x )
f 0
(x)
f (x) °1 = x°1
ex°x )
f 0
(x)
f (x) = 1+ x°1
ex°x )
f 0
(x)
f (x) = ex
°x+x°1
ex°x )
)
f 0
(x)
f (x) = ex
°1
ex°x ) (ln f (x))0
= (ln(ex
° x))0
ln f (x) = ln(ex
° x)+ c.
x = 0: ln f (0) = ln(e0
°0)+ c ) ln1 = ln1+ c ) c = 0
ln f (x) = ln(ex
° x).
µ y = lnx . "1 ° 1" µ
f (x) = ex
° x, x 2 R.
2. µ µ : f (x)+ f (°x) ∏ 2.
µ µ :
ex
° x+ e°x
+ x ∏ 2 , ex
+ e°x
∏ 2 , ex
+ 1
ex ∏ 2 , (ex
)2
+1 ∏ 2ex
,
, (ex
)2
°2ex
+1 ∏ 0 , (ex
°1)2
∏ 0,
. µ . " = "
ex
°1 = 0 , x = 0
3. f 0
(x) = (ex
° x)0
= ex
°1
f 0
(x) ∏ 0 , ex
∏ 1 , ex
∏ e0
, x ∏ 0
µ µ :
, µ (°1,0] f , µ
[0,+1) f .
2
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 3 of 17

4. µ x0 = 0 f µ µ
min f (x) = f (0) = 1
lim
x!°1
f (x) = lim
x!°1
(ex
° x) = +1
lim
x!+1
f (x) = lim
x!+1
(ex
° x)
(1°1)
= lim
x!+1
£
ex
(1° x
ex )
§
lim
x!+1
ex
= +1 lim
x!+1
x
ex
°0
0
¢
=
DLH
lim
x!1
1
ex = 0
lim
x!+1
°
1° x
ex
¢
= 1 µ lim
x!+1
f (x) = +1.
, f (A) = [1,+1).
4. µ ex
= x+ a , ex
° x = a , f (x) = a
i) Æ 2 (°1,1),
ii) a = 1, µ
iii) Æ 2 (1,+1),
5. f 0
(x) = ex
°1
f 00
(x) = (ex
°1)0
= ex
> 0, x 2 R
f µ µ .
:
3
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 4 of 17

5. 6. M(x1, f (x1)) µ µ (") µ
f (x), . (") :
y° f (x1) = f 0
(x1)(x° x1) )
y°(ex1 ° x1) = (ex1 °1)(x° x1)
µ x = 0 y = 0 :
0° ex1 + x1 = (ex1 °1)(0° x1) )
°ex1 + x1 = °x1ex1 + x1 )
x1ex1 ° ex1 = 0 )
ex1 (x1 °1) = 0 )
x1 = 1
µ M (1, e °1). µ :
y°(e °1) = (e °1)(x°1) )
y° e +1 = (e °1)x° e +1 )
y = (e °1)x
4
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 5 of 17

6. 7. f [0,1]
f (x) ∏ (e °1)x, :
E =
1R
0
[f (x)°(e °1)x]dx =
1R
0
(ex
° x°(e °1)x)dx =
1R
0
ex
dx°
1R
0
xdx°(e °1)
1R
0
xdx =
= [ex
]1
0 °
h
x2
2
i1
0
°(e °1)
h
x2
2
i1
0
= e °1°
°1
2 °0
¢
°(e °1)
°1
2 °0
¢
= e °1° 1
2 ° e°1
2 =
= e °1° 1
2 ° e
2 + 1
2 = e
2 °1 = e°2
2 ø.µ.
8. µ µ ex
∏ ex, x 2 R µ :
ex
∏ ex , ex
° x ∏ ex , f (x) ∏ (e °1)x, µ f
µ .
5
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 6 of 17

7. 3
f (x) = ln
≥
x+
p
x2 +1
¥
.
1. µ f .
2. f µµ
.
3. µ f µ .
4. µ f µ -
µ .
5. f f °1
(x) µ -
µ , µ µ µ .
6. µ (") Cf µ µ
µ µ x0 = 0 µ
Cf , µ (") x = 1.
7. E1 µ Cf , x0
x
x = 0,x = 1 E2 µ
Cf °1 , y0
y y = 0,y = 1 :
E1 + E2 = 2
≥
1°
p
2+ln(1+
p
2)
¥
8. :
lim
x!0
µ
f °1
(x)·
ex
x
∂
6
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 7 of 17

8. 4
1. x+
p
x2 +1 > 0
• x < 0, x+
p
x2 +1 > 0 )
p
x2 +1 > °x )
≥p
x2 +1
¥2
> (°x)2
)
x2
+1 > x2
+1 ) 1 > 0,
• x > 0, x+
p
x2 +1 > 0 )
p
x2 +1 > °x,
Af = R
2. µ Af = R x 2 R °x 2 R µ :
f (°x) = ln
≥
°x+
p
(°x)2
+1
¥
= ln
≥p
x2 +1° x
¥
= ln
≥p
x2+1°x
¥≥p
x2+1+x
¥
≥p
x2+1+x
¥ =
= ln
≥p
x2+1
¥2
°x2
≥p
x2+1+x
¥ = ln x2
+1°x2
≥p
x2+1+x
¥ = ln 1≥p
x2+1+x
¥ = ln1°ln
≥p
x2 +1+ x
¥
=
= °ln
≥p
x2 +1+ x
¥
= °f (x)
f Cf µµ
.
3. f
f 0
(x) =
≥
x+
p
x2+1
¥0
x+
p
x2+1
=
1+
≥p
x2+1
¥0
x+
p
x2+1
=
1+
(x2+1)
0
2
p
x2+1
x+
p
x2+1
=
1+ 62x
62
p
x2+1
x+
p
x2+1
=
=
x+
p
x2+1p
x2+1
x+
p
x2+1
1
= x+
p
x2+1≥
x+
p
x2+1
¥p
x2+1
= 1p
x2+1
f 0
(x) = 1p
x2+1
> 0, x 2 R, f R
.
4. f 00
(x) =
≥
1p
x2+1
¥0
= °
≥p
x2+1
¥0
≥p
x2+1
¥2 = °
62x
62
p
x2+1
x2+1
= ° x≥p
x2+1
¥
(x2+1)
f 00
(x) = 0 , ° x≥p
x2+1
¥
(x2+1)
= 0 , x = 0
f 00
(x) > 0 , ° x≥p
x2+1
¥
(x2+1)
> 0 , x < 0
f 00
(x) < 0 , ° x≥p
x2+1
¥
(x2+1)
< 0 , x > 0
7
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 8 of 17

9. :
µ (°1,0] Cf
µ [0,+1) Cf
µ x0 = 0 f µ O(0,0).
5. µ f 0
(x) = 1p
x2+1
> 0 f
R, "1°1", .
f (x) = y , ln
≥
x+
p
x2 +1
¥
= y , x+
p
x2 +1 = ey
,
p
x2 +1 = ey
° x
≥p
x2 +1
¥2
= (ey
° x)2
, x2
+1 = e2y
°2xey
+ x2
, 2xey
= e2y
°1 , x = e2y
°1
2ey
) f °1
(y) = e2y
°1
2ey = e2y
2ey ° 1
2ey = ey
2 ° 1
2ey = ey
2 ° e°y
2 ) f °1
(y) = 1
2 (ey
° e°y
)
, f °1
(x) = 1
2 (ex
° e°x
),x 2 R
g(x) = f °1
(x) = 1
2 (ex
° e°x
)
g0
(x) = 1
2 (ex
+ e°x
) > 0
g00
(x) = 1
2 (ex
° e°x
)
g00
(x) = 0 , 1
2 (ex
° e°x
) = 0 , ex
° e°x
= 0 , ex
= e°x
, x = 0
8
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 9 of 17

10. :
µ (°1,0] Cg
µ [0,+1) Cg
µ x0 = 0 g µ O(0,0).
6. µ (") Cf x0 = 0 :
9
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 10 of 17

11. (" :) y° f (0) = f 0
(0)(x°0) )
y°0 = 1·(x°0) )
y = x
Cf [0,+1) f (x) ∑ x, x 2 [0,+1)
" = " µ x = 0 f (x)° x ∑ 0.
:
E =
1R
0
|f (x)° x|dx =
1R
0
(x° f (x))dx =
1R
0
xdx°
1R
0
f (x)dx =
=
h
x2
2
i1
0
°
1R
0
ln
≥
x+
p
x2 +1
¥
dx = 1
2 °
1R
0
ln
≥
x+
p
x2 +1
¥
dx
I1 =
1R
0
ln
≥
x+
p
x2 +1
¥
dx =
1R
0
(x)0
·ln
≥
x+
p
x2 +1
¥
dx =
=
h
x·ln
≥
x+
p
x2 +1
¥i1
0
°
1R
0
x· xp
x2+1
dx =
h
x·ln
≥
x+
p
x2 +1
¥i1
0
°
1R
0
x· xp
x2+1
dx =
= ln(1+
p
2)°
1R
0
x· xp
x2+1
dx
p
x2 +1 = t ) x2
+1 = t2
) 2xdx = 2tdt ) xdx = tdt
x = 0 ! t = 1 x = 1 ! t =
p
2
I1 = ln(1+
p
2)°
p
2R
1
t
t dt = ln(1+
p
2)°[t]
p
2
1 = ln(1+
p
2)°
p
2+1
E = 1
2 °ln(1+
p
2)+
p
2°1 =
p
2° 1
2 °ln(1+
p
2).
7. µµ Cf Cg y = x : E1 = E2
:
E1 + E2 = (ABO°)°2E = (OB)·(AB)°2
°p
2° 1
2 °ln(1+
p
2)
¢
=
= 1·1°2
p
2+1+2ln(1+
p
2) = 2°2
p
2+2ln(1+
p
2) = 2·
°
1°
p
2+ln(1+
p
2)
¢
8. lim
x!0
≥
f °1
(x)· ex
x
¥
= lim
x!0
≥
ex
°e°x
2 · ex
x
¥
= lim
x!0
e2x
°1
2x
°0
0
¢
=
DLH
lim
x!0
(e2x
°1)
0
(2x)0 =
= lim
x!0
62e2x
62 = lim
x!0
e2x
= 1
10
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 11 of 17

12. 5
f : R ! R µ
:
°
x2
+1
¢
·
h
f (x)°ln
p
x2 +1
i
= x·
£
x°
°
x2
+1
¢
f 0
(x)
§
, 8x 2 R (3)
1. f (x).
2. .
3. µ µ .
4. µ µ µ .
5. µ
.
6. 2f (x) = x+ln 2
e , x 2 [1,+1)
7. E2
E1
= ln4
4°º
°1, :
• E1 µ
f x = °1, x = 1 y = ln
p
2
2
• E2 µ
f x = °1, x = 1 y = 0.
8. lim
x!+1
f (x)+x
f (x)+ex .
11
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 12 of 17

13. 6
1. 3 µ :
°
x2
+1
¢
· f (x)°
°
x2
+1
¢
·ln
p
x2 +1 = x2
° x
°
x2
+1
¢
f 0
(x) ,
°
x2
+1
¢
· f (x)+ x
°
x2
+1
¢
f 0
(x) =
°
x2
+1
¢
·ln
p
x2 +1+ x2 :(x2
+1)
,
f (x)+ xf 0
(x) = ln
p
x2 +1+ x· x
x2+1
,
(x· f (x))0
= (x)0
·ln
p
x2 +1+ x·
≥
ln
p
x2 +1
¥0
,
(x· f (x))0
=
≥
x·ln
p
x2 +1
¥0
x· f (x) = x·ln
p
x2 +1+ c
x = 0: 0 = 0+ c , c = 0
µ , x· f (x) = x·ln
p
x2 +1+ c
: x6=0
, f (x) = ln
p
x2 +1, x 6= 0
f (x) = ln
p
x2 +1, x 6= 0
f µ µ , µ x0 = 0,
µ : f (0) = lim
x!0
f (x) = lim
x!0
≥
ln
p
x2 +1
¥
= ln1 = 0 , f (0) = 0,
f (x) = ln
p
x2 +1.
, f (x) = ln
p
x2 +1, x 2 R .
2. Af = R, x 2 R °x 2 R µ
f (°x) = ln
p
(°x)2
+1 = ln
p
x2 +1 = f (x), f .
3. x 2 R :
f 0
(x) =
≥
ln
p
x2 +1
¥0
=
≥p
x2+1
¥0
p
x2+1
=
2x
2
p
x2+1
p
x2+1
= x
x2+1
f 0
(x) = 0 , x
x2+1
= 0 , x = 0
f 0
(x) > 0 , x
x2+1
> 0
x2
+1>0
, x > 0
µ :
12
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 13 of 17

14. , µ (°1,0] f , µ
[0,+1) f .
µ x0 = 0 f µ µ
min f (x) = f (0) = 1
4. f 00
(x) =
≥
x
x2+1
¥0
=
(x)0
·x2
+1°x(x2
+1)
0
(x2+1)2 = x2
+1°2x2
(x2+1)2 = °x2
+1
(x2+1)2
f 00
(x) = 0 , °x2
+1
(x2+1)2 = 0 , x = °1 x = 1
µ :
µ (°1, ° 1], [1 , +1) f , µ
[°1,1] f .
µ A
°
°1,ln
p
2
¢
B
°
1,ln
p
2
¢
-
µ µ .
5. [Af = R µ .
µ :
∏ = lim
x!+1
f (x)
x = lim
x!+1
ln
p
x2+1
x
(1
1 )
= lim
x!+1
x
x2+1
1 = lim
x!+1
x
x2+1
= 0
Ø = lim
x!+1
f (x) = lim
x!+1
≥
ln
p
x2 +1
¥
= +1
µ µ +1, µ °1.
µ :
13
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 14 of 17

15. 6. x 2 [1,+1) :
2f (x) = x+ln 2
e ,
2f (x) = x+ln2°ln e ,
f (x) = 1
2 x+ 1
2 ln2° 1
2 ,
f (x) = 1
2 x° 1
2 + 1
2 ln2
µ f B
°
1,ln
p
2
¢
-
:
(") : y° 1
2 ln2 = f 0
(1)(x°1) ,
y° 1
2 ln2 = 1
2(x°1) ,
y = 1
2 x° 1
2 + 1
2 ln2
f x 2 [1,+1) :
f (x) ∑ 1
2 x ° 1
2 + 1
2 ln2 " = " µ µ µ
x = 1.
: f (x) = 1
2 x° 1
2 + 1
2 ln2 , x = 1
14
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 15 of 17

16. 7. µ E2
x 2 [°1,1] f (x) ∏ 0 :
E2 =
R1
°1 f (x)dx =
R1
°1 f (x)dx =
R1
°1 ln
p
x2 +1dx
f :ÆΩø∂Æ
= 2·
R1
0 ln
p
x2 +1dx =
= 2·
R1
0 (x)0
ln
p
x2 +1dx =2
µh
x·ln
p
x2 +1
i1
0
°
R1
0 x·
≥
ln
p
x2 +1
¥0
dx
∂
=
= 2
≥
ln
p
2°0°
R1
0 x· x
x2+1
dx
¥
= 2ln
p
2°2·
R1
0
x2
x2+1
dx =
= 2· 1
2 ln2°2
R1
0
x2
+1°1
x2+1
dx = ln2°2
≥R1
0
x2
+1
x2+1
dx°
R1
0
1
x2+1
dx
¥
=
= ln2°2+2
R1
0
1
x2+1
dx = ln2°2+2I1
I1 =
R1
0
1
x2+1
dx
x = "'t ) dx = 1
æ¿∫2t
dt = (1+"'2
t)dt
x = 0 : "'t = 0 , t = 0
x = 1 : "'t = 1 , t = º
4
I1 =
Rº
4
0
1
1+"'2t
(1+"'2
t)dt =
Rº
4
0 dt = º
4
µ , E2 = ln2°2+2· º
4 = ln2°2+ º
2 ø.µ.
µ E1 : E1 = (AB°¢)°E2 = 2· 1
2 ·ln2°
°
ln2°2+ º
2
¢
=
2° º
2 ø.µ.
,
E2
E1
=
ln2°2+º
2
2°º
2
=
ln2°(2°º
2 )
2°º
2
= ln2
2°º
2
°1 = 2ln2
4°º
°1 = ln4
4°º
°1 .
15
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 16 of 17

17. 8. lim
x!+1
f (x)+x
f (x)+ex = lim
x!+1
ln
p
x2+1+x
ln
p
x2+1+ex
(1
1 )
= lim
x!+1
≥
ln
p
x2+1+x
¥0
≥
ln
p
x2+1+ex
¥0 =
= lim
x!+1
x
x2+1
+1
x
x2+1
+ex = lim
x!+1
x+x2+1
x2+1
x+(x2+1)ex
x2+1
= lim
x!+1
x2
+x+1
x+(x2+1)ex
(1
1 )
=
= lim
x!+1
2x+1
2xex+(x2+1)ex+1
(1
1 )
= lim
x!+1
2
ex
(x2+1+4x+2)
= lim
x!+1
2
ex
(x2+4x+3)
= 0
16
23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 17 of 17