Trabalho de Matemática - Funções

Matemática - Funções
(Mack-SP) Na função de IR→IR, definida por f ( x ) =3 x+ 1 , calcule

f ( 235 ) − f ( 129 )
.
106

Resolução: f ( x ) =3 x+ 1→ f ( 235 )= 3 x +1 → f ( 235 ) =3.235+1 → f ( 235 ) =705+1 → f ( 235 ) =706 f ( x ) =3 x+ 1→ f ( 129 )=3 x +1 → f ( 129 ) =3.129+1→ f ( 129 )= 387+1 → f ( 129 ) =388 f ( 235 ) − f ( 129 ) 706 −388 318
→
→
→3
106
106
106
(PUC) O gráfico da função quadrática f ( x ) = x ²+ax +3 passa pelo ponto P(1,2). Determine o valor de a .
Resolução:
P ( 1,2 ) → P ( x , y )

f ( x)= y

f ( x ) = x ²+ax +3 → f ( 1 ) =1²+a .1+3 → 2= 1+ a+3 →2= 4+ a→ a= 2− 4 → a=− 2
(Fuvest-SP) Se f ( x ) =

1
4
, quanto vale f ( √
7) ? x ²+ 1

Resolução:
1
4
1
4
1
√7 − 1 → f ( √4 7 )= √7 −1 → f ( √4 7 )= √7 − 1 f ( √7 ) = 4 2 → f ( √7 )=
→ f ( √ 7 )=
.
6
√7 +1
√7+1 √7 − 1
√49 −1
√7 +1
4

1
(PUCCAMP-SP) Dadas as funções reais f ( x ) = −1, x ≠ 1 e g ( x )= 2 x −4 , o valor de x ( f ∘ g )( 2 ) + ( g ∘ f ) 1 é igual a:
2
a) 7
b)0
c)-9
d)-7
e)n.d.a.
1
1
1
f ( g ( x ) )=
→ f ( g ( 2) ) =
→ f ( g ( 2) )=
→ f ( g ( 2 )) =− 1
−1
( 2 x − 4 ) −1
( 2.2− 4 ) −1
1
1
1
1
1
1
2
g ( f ( x )) = 2.
−4→g f
=2.
−4→g f
=2.
−4→g f
=
−4→ x−1 2
1
2
−1
2
−1
−1
2
2
2
1
2
1
4
1
1
4+ 4
1
8
1
= 2.
−4→ g f
=
−4 → g f
=g f
=
→g f
=
→g f
2
−1
2
−1
2
2
−1
2
−1
2

()

( ( ))
( ( ))
( ( ))
( ( )) ( ( ))

( )

( ( ))

g f

()

()

( ( ))
( ( )) ( ( ))
()

( f ∘ g )( 2 ) + ( g ∘ f ) 1 =− 1+ ( − 8 ) → ( f ∘ g )( 2 )+( g ∘ f ) 1 =−1 −8 → ( f ∘ g )( 2 )+ ( g ∘ f ) 1 = −9
2

Resposta c

2

2

=− 8