Princippio de integrales

Determinar la ecuación de la recta tangente a la gráfica de f en un punto dado. Utilizar aproximación lineal para completar la gráfica.

1)

T(x) = 4x-4


1.9
1.99
2
2.01
2.1

3.613.9601
4
4.0401
4.41

3.6
3.96
4
4.04
4.40

F(x) =, (2,4)
F’(x) = 2x
Mtg = f’ (2) = 2(2) = 4


Y – 4 = 4 (x-2)
Y= 4x – 8+4
Y= 4x - 4

f(1.9)=4(1.9) – 4 = 3.6 f(1.99)= 4(1.99) - 4= 3.96
f(2)=4(2) – 4 = 4 f(2.01) = 4(2.01) – 4 = 4.04
f(2.1) = 4(2.1) – 4 = 4.4

3)

T(x) = 80x-128


1.9
1.99
2
2.01
2.124.79099
31.2079
32
32.8080
40.8410

24
31.2
32
32.8
40.0
F(x) =, (2,32)
F’(x) = 5
Mtg = f’ (2) = 5 = 80


Y - 32 = 80(x – 2)
Y = 80x – 160 + 32
Y = 80x – 128

f(1.9) = 80(1.9) – 128= 24 f(1.99) = 80(1.99) – 128 = 31.2
f(2) = 80(2) -128 = 32 f(2.01) = 80(2.01) – 128 = 32.8
f(2.1) = 80(2.1) – 128 = 40

5)T(x) = cos2(x-2)+sen2


1.9
1.99
2
2.01
2.1

0.94
0.91
0.909
0.905
0.863

0.951
0.913
0.909
0.905
0.868

F’(x)= cos(x)
y-sen(2) = cos(x)(x-2)
y = cos(x)(x-2)+sen(x)

f’(1.9) =cos(1.9)(1.9-2) + sen(2) = 0.94
f’(1.99) = cos(1.99)(1.99-2) + sen(2) = 0.913
f’(2) = cos(2)(2-2) + sen(2) = 0.909
f’(2.01) = cos(2.01)(2.01-2) + sen(2) = 0.905
f’(2.1) = cos(2.1)(2.1-2) + sen(2)= 0.858


En los ejercicios 21 a 24 emplear diferenciales y grafica de f para aproximar a) f(1.9) y b) f(2.04).
21. - dy=mtg=
f(1.9)= f(2)+dy
=f(2)+f’(2)dx=f(2)+f’(2)(1.9-2)
=1+(1) (-0.1)
R =.09
f(2.04) =f(2)+dy
=f(2)+f’(2)dx
=f(2)+f’(2)(2.04-2)
=1+(1) (0.04)
R =1.04

23.-dy= mtg=
f(1.9)= f(2)+dy
=f(2)+f’(2)dx
=f(2)+f’(2)(1.9-2)
=1+(-1/2) (-0.1)
R =1.05
f(2.04) =f(2)+dy
=f(2)+f’(2)dx...