Actividad 1 Calculo Diferencial.
Actividad 1 Calculo Diferencial.
Actividad 1 Calculo Diferencial.
CALCULO DIFERENCIAL
IBAGUÉ – TOLIMA
2021
1. Dada la ecuación F ( x )=3 x +2
F (1) = 3+2
F (1) = 5
F (-2) = -6 + 2
F (1) = -4
F (-2) = 3X^2 + 2
F (x+h) = 3x+3h+2
2.
Dada la ecuación F ( x )=−2 x +5
F (3) = -6 + 5
F (3) = -1
F (-1) = 2 + 5
F (-1) = 7
Calcule: F (x + h) = -2*(x + h) + 5
F (x + h) = -2x-2h+ 5
3. Dada la ecuación F ( t )=5 t+ 7
F (1) = 5 + 7
F (1) = 12
F (-3) = -15 + 7
F (-3) = -8
F (c) = 5c + 7
F (1) + F (c) = 5 + 7 + 5c + 7
F (1) + F (c) = 5c + 19
4.
Dada la ecuación F ( x )=−4 x +3
F (a) = -4a + 3
Calcule: F (a + 1) = -4* (a + 1) + 3
F (a + 1) = -4a – 4 + 3
F (a + 1) = -4a -1
F (3) = 9
F (-2) = 4
F (a) = a^2
2
Calcule: F ( √ x )=( √ x ) =x
Calcule: F (x + h) = (x + h) ^2
F (x + h) = (x + h) ^2
6.
Dada la ecuación F ( x )=3 x 2+ 7
Calcule: F (c + h) = 3*(c + h) ^2 + 7
F (c + h) = 3*c^2 + h^2 + 7
F (c + h) = 3c^2 + 3h^2 + 7
F (c + h) – F (c)= 3h^2 + 14
(1.30) Evalué los siguientes limites:
2
1. lim 3 x +7 x−¿1
x →2
2
¿ 3 ( 2 ) + 7 ( 2 )−1
¿ 12+14−1
=25
3. lim ¿ x+ 1
x →3 x−2
3+1
= 3−2
4
= 1
lim ¿ x 2−25
5. x →5
√ x 2 +11
lim ( x−5 )( x +5 )
x→ 5
√5 2+1 1
( 5−5 ) ( 5+5 )
= 6
10
=6
5
=3
7. lim ¿ x2 −4
x→−2
x2 +3 x +2
(−2 )2 −4
= 2
(−2 ) + 3 ( 2 ) +2
4−4
= 4−6+ 0
0
=0
x−2
lim
x →−2 x+1
−2−2 −4
= −2+1 = −1 =4
lim x 2−5 x+ 6
9 . x→ 3
x−3
( 3 )2−5 ( 3 ) +6
= 3−3
9−15+ 6 0
= 0
=
0
( x−3 )( x−2 )
=lim
x →3 ( x−3 )
lim X−2
x →3
3−2=1
lim x 2 +3 x+ 2
11 . x →−1
x 2−1
2
13.xlim
→−2
x +4 x+ 4
x 2−4
(1 )2 +1−2
= 2
( 1 ) −3 ( 1 )+2
1+ 1−2 0
= 1−3+2 = 0
lim ( x +2 ) ( x−1 )
x→ 1
( x−2 ) ( x−1 )
lim 1+2
x→ 1 3
= =−3
1−2 −1
lim 9−x
x →17.
9
√ x−3
9−9 0
¿ =
√ 9−3 0
lim 9−x
x+ 3
∙√
x→9
√ x−3 √ x+ 3
lim ( 9−x ) ( √3+3 )
x→9
( √ x−3 )( √ x+ 3 )
lim 9− χ ⋅ √ x+ 3
x →9
x−9
¿ √ 9+3
=3+3
=6
lim x 3−1
19.
x→ 1
2
x −1
( 1 )3 −1 1−1 0
= ( 1 )2−1 = 1−1 = 0
lim ( x−1 ) ( x 2+ x +12 )
x→ 1
( x−1 ) ( x+ 1 )
¿ √ 4−2 = 2−2 = 0
3
( 4 ) −64 64−64 0
lim √ x−2
x +2
x→4
3
∙√
x −64 √ x +2
lim ( √ x−2 ) ( √ x+2 )
x→4
( x 3−64 ) ( √ 4 +2 )
lim x−4
x→ 4
3
( x −64 ) ( √ 4+2 )
lim x −4
x→ 4
( x−4 ) ( x + 4 x+ 4 2) ( √ 4 +2 )
2
1
= ( 16+16+16 )( 2+2 )
1
= 96
25.lim 4 + x−2
√
x→0
x
lim √ x+ 4−2
x +4 +2
x→0
x√
2 √ x +4 +2
lim x
x →0
√ x + 4+2
1
=4
x+ 3−2
27 ⋅ lim ¿ √
x →1 x 2−1
x+ 3−2 √ x +3+2
lim ¿ √
x →1 x 2−1
. ( √ x +3+2 )
2
(−2 )2−( √ x+3 )
lim ¿
x →1 ( x−1 ) ( x+1 ) ( √ x+3+ 2 )
4−x−3
lim ¿
x →1 ( x +1 )( x−1 ) l √ x +3+2
(−x+1 ) ⋅ (−1 )
lim ¿
x →1 ( x−1 ) ( x+1 ) ( √ x+3+ 2 ) (−1 )
1
=8