# Exercicio do livro calculo a

Páginas: 17 (4238 palavras) Publicado: 12 de abril de 2013
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CAPÍTULO 2
SEÇÃO 2.10 – página 20
1. Se f ( x ) = x2 − 4 , achar: x −1

a) f (0 ) =

02 − 4 − 4 = = 4. 0 −1 −1

b) f (− 2 ) =

(− 2)2 − 4 = 4 − 4 (− 2) − 1 − 3
2

= 0.

1 1   −4 −4 1 − 4t 2 t 1 − 4t 2 t  t2 = = ⋅ = . c) f (1 t ) = 1 1− t 1− t t2 t −t2 −1 t t d)

( x − 2 )2 − 4 = x 2 − 4 x + 4 − 4 = f (x − 2) =
x − 2 −1
2

x −3

x 2 − 4x . x−3

1 1   −4 −41 − 16 2 − 15 15 2 4 e) f (1 2) = = = ⋅ = = . 1 1 4 1− 2 − 2 2 −1 −1 2 2
f)

(t ) − 4 = t f (t ) = t −1 t
2 2 2 2

4 2

−4 . −1

2. Se f ( x ) =

3x − 1 , determine: x−7

a)

5 f (− 1) − 2 f (0 ) + 3 f (5) 7

f (− 1) =

3(− 1) − 1 − 3 − 1 − 4 1 = = = −1− 7 −8 −8 2

f (0 ) =

3× 0 −1 −1 1 = = 0−7 −7 7

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f (5) =

3(5) − 1 15 − 1 14 = = = −7 5−7 −2 −2

Portanto,5 f (− 1) − 2 f (0 ) + 3 f (5) = 7

5 =

= = =

1 1 − 2 + 3 (− 7 ) 2 7 7 5 2 − − 21 2 7 7 35 − 4 − 294 1 ⋅ 14 7 − 263 1 − 263 ⋅ = 14 7 98

b)

[ f (− 1 2)]2

 −1  −1 3  = 2  − 1 −7     2 
2

2

 −3−2  2    2  = −5 ⋅ 2  = 1   9  − 1 − 14   2 − 15     2  3(3 x − 2 ) − 1 9 x − 7 = . 3x − 2 − 7 3x − 9

c) f (3 x − 2 ) =

d) f (t ) + f (4 t ) =

43  − 1 3t − 1  t  3t − 1 12 − t t = + = + ⋅ 4 t −7 t −7 t 4 − 7t −7 t (3t − 1) (4 − 7t ) + (12 − t ) (t − 7 ) = 12t − 21t 2 − 4 + 7t + 12t − 84 − t 2 + 7t = 4t − 7t 2 − 28 + 49 − 7t 2 + 53t − 28 − 22t 2 + 38t − 88 = . − 7t 2 + 53t − 28

30

e)

f (h ) − f (0 ) = h
 3h − 1 3 ⋅ 0 − 1  1 = − ⋅ 0−7  h  h−7  3h − 1 1  1 = − ⋅  h−7 7 h 21h − 7 − 1(h − 7 ) 1 = ⋅ 7(h − 7 ) h 20h 1= ⋅ 7(h − 7 ) h 20 = 7(h − 7 )

f) f [ f (5)]

f (5) =

3 ⋅ 5 − 1 14 = −7 = 5−7 −2
3(− 7 ) − 1 − 21 − 1 − 22 11 = = = . (− 7 ) − 7 − 14 − 14 7

f [ f (5)] = f (− 7 ) =

3. Dada a função f ( x ) = x − 2 x , calcular. f (− 1) , f (1 2 ) e f (− 2 3) . Mostrar que

f (a ) = − a .

f (− 1) = − 1 − 2(− 1) = 1 + 2 = 3
1 1 1− 2 −1 = −1= = . 2 2 2 2 − 2 2 4 6 f (− 2 3) = − 2 3 − 2  = += = 2.  3  3 3 3 f (1 2 ) = 1 2 − 2

f (a ) = a − 2 a = a −2 a = −a

4. Se f ( x ) =

ax + b e d = − a , mostre que f ( f ( x )) = x cx + d

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ax + b cx − a

f (x ) =

 ax + b  f ( f ( x )) = f    cx − a   ax + b  a⋅ +b  cx − a  =  ax + b  c⋅  + (− a )  cx − a  a (ax + b ) +b = cx − a c (ax + b ) −a cx − a a (ax + b ) + b(cx − a ) cx − a = ⋅ cx − a c(ax + b ) −a(cx − a ) a 2 x + ab + bcx − ab cax + cb − acx + a 2 a 2 x + bcx x a 2 + bc =x = = 2 cb + a 2 a + bc =

(

)

5. Se f ( x ) = x 2 + 2 x , achar
geometricamente.

f (a + h ) − f (a ) , h ≠ 0 e interpretar o resultado h

f (a + h ) − f (a ) (a + h ) + 2(a + h ) − a 2 + 2a = h h 2 2 a + 2ah + h + 2a + 2h − a 2 − 2a h(2a + h + 2 ) = = h h = 2a + 2 + h A Figura que segue mostra ainterpretação geométrica. Nesta figura, α é o ângulo formado pela reta que passa pelos pontos (a, f (a )) e (a + h, f (a + h )) e o eixo positivo dos x . O quociente obtido representa a tangente do ângulo α .

2

(

)

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6. Dada Φ( x ) =

x −1 . Forme as expressões Φ(1 x ) e 1 Φ( x ) . 2x + 7

1− x 1 x −1 1− x x 1− x Φ(1 x ) = = x = . ⋅ = 1 2 + 7x x 2 + 7x 2 + 7x 2⋅ + 7 x x

1 1 2x + 7 = = .x −1 Φ( x) x −1 2x + 7

7. Dada a função f ( x ) = x 2 + 1 , mostrar que para a ≠ 0 f (1 a ) = f (a ) a 2 .

1 1+ a2 f (a ) 1 f (1 a ) =   + 1 = 2 + 1 = = 2 , para a ≠ 0 . 2 a a a a

2

8. Dada a função f ( x ) =

1 h , mostrar que f (1 + h ) − f (1) = − . Calcular. x 1+ h

33

f (a + h ) − f (a ) .
f (1 + h ) − f (1) = 1 1 1−1− h − h − = = 1+ h 1 1+ h 1+ h

f (a + h ) − f(a ) =

1 1 a−a−h −h − = = a + h a a (a + h ) a (a + h )

9. Seja f (n ) a soma dos n termos de uma progressão aritmética. Demonstrar que f (n + 3) − 3 f (n + 2 ) + 3 f (n + 1) − f (n ) = 0 . f (n ) = a1 + a1 + r + a1 + 2r + L + a1 + (n − 1)r = na1 + (1 + 2 + 3 + L + (n − 1) )r f (n + 3) = (n + 3)a1 + [1 + 2 + L + (n + 2 )] r f (n + 2 ) = (n + 2 )a1 + [1 + 2 + L + (n + 1)] r f (n + 1) =...

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