uba 21 modelo parcial

1. Si f (x) =

x−1
y f (0) = 1 ent´onces ’a’ es igual a: −6; −3; 0; 3; 6
ax + 3
f (x) =

x−1
ax + 3

(ax + 3) − a(x − 1)
3+a
=
2
(ax + 3)
(ax + 3)2
3+a
f (0) = 1 ⇔
=1
(a(0) + 3)2
3+a
=1
f (0) = 1 ⇔9
f (0) = 1 ⇔ 3 + a = 9
=

f (0) = 1 ⇔ a = 6
Comprobaci´on:
f (0) =

9
3 + (6)
=1
2 =
9
[(6(0)) + 3]
1

2. La derivada de la funci´on f es f (x) = (3x + 1) 2 y adem´as es f (0) = 1.
F (x) =

f (x) dx=

Sustituci´on, u = 3x + 1; du = 3dx ent´onces dx =
1

(3x + 1) 2 dx =

1

(u) 2

1

(3x + 1) 2 dx
du
. Reemplazamos:
3

3
3
2
du
1 2
= . (u) 2 + C = (3x + 1) 2 + C
3
3 3
9

Averiguamos C con f (0) =1:
3
2
(3(0) + 1) 2 + C = 1
9
2
f (0) = 1 ⇔ · 1 + C = 1
9
2
f (0) = 1 ⇔ C = 1 −
9
7
f (0) = 1 ⇔ C =
9
3
2
7
∴ F (x) = (3x + 1) 2 +
Opci´on 3
9
9

f (0) = 1 ⇔

3. El a´rea de la regi´on determinada porlas curvas y = x3 e y = 4x:
Datos:
f (x) = x3
g(x) = 4x
Intersecci´on, f (x) = g(x):
x3 = 4x
x3 − 4x = 0
x(x2 − 4) = 0
x(x + 2)(x − 2) = 0
⇒ x = 0 ∨ x = −2 ∨ x = 2
1

El signo de los intervalos:(−∞; −2)(−2; 0)(0; 2)(2; +∞) nos interesa desde −2 hasta 2.
(−2; 0) : f (−1) = −1 ∧ g(−1) = −4 ∴ f (x) > g(x)
(0; 2) : f (1) = 1 ∧ g(1) = 4 ∴ f (x) < g(x)
0

2

(f (x) − g(x))dx +

A(r) = F (x) =
−2
0

=(g(x) − f (x))dx =
0

2

(x3 − 4x)dx +

−2

(4x − x3 )dx Opci´
on 3

0

4. El ´area de regi´on determinada por las curvas y = x2 + 9; y = −x2 + 9 y la recta x = 4.
Datos:
f (x) = x2 + 9
g(x) = −x2 + 9Directamente el signo con x = 4 (L´ımite superior):
f (4) = (4)2 + 9 = 25
g(4) = −(4)2 + 9 = 5
f (4) > g(4)

∴
Interseccion:

x2 + 9 = −x2 + 9
2x2 = 0 ⇒ x = 0 L´
ımite inferior
´
Area
de la regi´on:
4A(r) = F (x) =

4

(f (x)−g(x))dx =
0

0

0
4

··· =

4

2 3
128
2
x = (4)3 − 0 =
3 0 3
3

2x2 dx =

0

5. Si f (x) = k.sen(x) con k ∈ R, entonces f

π
2

= −4.

f (x) = kcos(x)
f (x) = k(−sen(x)) =−ksen(x)
π
π
f
= −4 ⇔ −ksen
= −4
2
2
π
π
f
= −4 ⇔ ksen
=4
2
2
π
f
= −4 ⇔ k · (1) = 4
2
π
f
= −4 ⇔ k = 4
2
Comprobaci´on:
1

f

π
2

4

(x2 +9−(−x2 +9))dx =

= −4 · sen
2

π
2

= −4

(x2 +9+x2...