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Prof. Roberto Mauro Moura
Função Afim matematicaaki-com.webnode.com
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introdução πππΓ‘πππ ππππππ=ππππ+π,ππ . (πππππ π
ππ ππππ
ππ ππ πΓͺπ)
Um representante comercial recebe, mensalmente, um salΓ‘rio composto de duas partes: uma parte fixa, no valor de R$ 1500,00 e uma parte variΓ‘vel, que corresponde a uma comissΓ£o de 6% (0,06) sobre o total das vendas que ele faz durante um mΓͺs. Nestas condiçáes, podemos dizer que: πππΓ‘πππ ππππππ=ππππ+π,ππ . (πππππ π
ππ ππππ
ππ ππ πΓͺπ) Podemos escrever essa função usando variΓ‘veis (letras minΓΊsculas do nosso alfabeto. πππΓ‘πππ ππππππ=π π ππ π π ππ π. πππππ π
ππ ππππ
ππ ππ πΓͺπ=π π π =ππππ+π,πππ ππ π π =ππππ+π,πππ ππ π=ππππ+π,πππ π¬πππ Γ© ππ πππππππ π
π πππçãπ ππππ.
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Definição de Função Afim
πππ ππ’πçãπ π:πΌπ
βπΌπ
πβπππβπ π ππ’πçãπ ππππ ππ’ππππ ππ₯ππ π‘ππ ππππ πΓΊπππππ πππππ π π π π‘ππ ππ’π π π =ππ+π ππππ π‘πππ π π π°πΉ. πΈπ₯ππππππ : π π₯ =2π₯+1 βπ=2 π π=1 π π₯ =βπ₯+4 βπ=β1 π π=4 π π₯ = 1 3 π₯β5 βπ= 1 3 π π=β5 π π₯ =4π₯ βπ=4 π π=0 π π₯ =β βπ=0 π π=β3
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Casos particulares importantes da função afim π π =ππ+π
1ΒΊ) função identidade π:πΌπ
βπΌπ
ππππππππ πππ π π =π ππππ π‘πππ π₯ π πΌπ
. πππ π π πππ π, π=1 π π=0. 2ΒΊ) função linear π:πΌπ
βπΌπ
ππππππππ πππ π π =ππ ππππ π‘πππ π₯ π πΌπ
. πππ π π πππ π, π=0. πΈπ₯ππππππ : π π₯ =β2π₯ π π₯ = 1 5 π₯ π π₯ = 3 π₯ π π₯ =0 ;ππ π‘π ππ’πçãπ Γ© πβπππππ πππππ‘πππππππ‘π ππ’ππ. 3ΒΊ) função constante π:πΌπ
βπΌπ
ππππππππ πππ π π =π ππππ π‘πππ π₯ π πΌπ
π=0. π΄πππ’ππ ππ₯ππππππ : π π₯ = 3 4 π π₯ =3 π π₯ =β2 π π₯ = 2 4ΒΊ) Translação (da função identidade) π:πΌπ
βπΌπ
ππππππππ πππ π π =π+π ππππ π‘πππ π₯ π πΌπ
π πβ 0. πππ π π πππ π, π=1. π π =π+π π π =π+ π π π π =πβ π π
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valor de uma Função Afim valor Inicial da Função Afim
π π£ππππ ππ π’ππ ππ’πçãπ ππππ π π₯ =ππ₯+π ππππ π₯= π₯ 0 Γ© ππππ πππ π π₯ 0 =π π₯ 0 +π. Exemplo: π π₯ =5π₯+1, determinar: π 1 =5.1+1βπ 1 =6 π β3 =5. β3 +1βπ β3 =β14 π = βπ =2 π π₯+β =5. π₯+β +1βπ π₯+β =5π₯+5β+1 valor Inicial da Função Afim ππ’ππ ππ’πçãπ ππππ π π₯ =ππ₯+π ππππ π=π 0 πβπππβπ π π£ππππ πππππππ ππ ππ’πçãπ π. Por exemplo: π π£ππππ πππππππ ππ ππ’πçãπ π π₯ =β2π₯+3 Γ© 3, ππππ π 0 =β2.0+3=3.
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Determinação de uma função afim π π =ππ+π
Uma função afim π π₯ =ππ₯+π fica inteiramente determinada quando conhecemos dois de seus valores π₯ 1 π π π₯ 2 ππππ ππ’πππ ππ’ππ π₯ 1 π π₯ 2 πππππ , πππ π₯ 1 β π₯ 2 . Por exemplo: πΈπ ππππ£π π ππ’πçãπ ππππ π π₯ =ππ₯+π, π ππππππ ππ’π π 2 =β2 π π 1 =1. 2π+π=β2 π+π=1 2π+π=β2 β2πβ2π=β2 π+π=1 π π =ππ+π .(βπ) π+4=1 π π =βππ+π βπ=β4 .(β1) π=1β4 π=4 π=β3 Taxa de Variação da função afim π π =ππ+π π πππΓ’πππ‘ππ π ππ π’ππ ππ’πçãπ ππππ π π₯ =ππ₯+π Γ© πβπππππ ππ π‘ππ₯π πππ£ππππçãπ ππ’ π‘ππ₯π ππ ππππ ππππππ‘π. ππππ πππ‘Γͺβπππ , πππ π‘ππ ππππ ππππ‘ππ ππ’πππ ππ’ππ, πππΓ©π πππ π‘πππ‘ππ . π₯ 1, π( π₯ 1 ) π π₯ 2 , π π₯ 2 , ππ π ππ π π₯ 1 =π π₯ 1 +π π π π₯ 2 =π π₯ 2 +π, πππ§ππππ π π₯ 2 -π π₯ 1 , π‘ππππ π π₯ 2 -π π₯ 1 =π π₯ 2 +π β π π₯ 1 +π βπ π₯ 2 βπ π₯ 1 =π π₯ 2 β π₯ 1 βπ= π π π βπ π π π π β π π π₯ 2 β π₯ 1 π π₯ =5π₯+2, ππππ π π‘ππ₯π ππ π£ππππçãπ Γ© π=5. π π₯ =β2π₯+3, ππππ π π‘ππ₯π ππ π£ππππçãπ Γ© π=β2.
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caracterização da Função Afim
πΏππππππππ ππ’π π’ππ ππ’πçãπ π:π΄βπΌπ
πππ π΄βπΌπ
Γ©: πͺππππππππ:π π π₯ 1 < π₯ 2 , πππ‘Γ£π π π₯ 1 <π π₯ 2 ; π«ππππππππππ:π π π₯ 1 < π₯ 2 , πππ‘Γ£π π π₯ 1 >π π₯ 2 . Γ possΓvel provar que: π·πππ π’ππ ππ’πçãπ π:πΌπ
βπΌπ
, πππππππππ ππ π
ππππππππππ, π π π πππππππΓ§π ππ π π+π βπ π ππππππππ ππππππ ππ π πππ πΓ£π ππ π, πππ‘Γ£π π Γ© π’ππ πππçãπ ππππ. Exemplo: π π₯ =3π₯β4 Γ© ππππ ππππ‘π π π π₯ 1 =2 π π₯ 2 =3 πππ 2<3 β΄ π 2 =2 π π 3 =5 Logo π 2 < π 3 π ππππ ππ’πππ ππ’ππ π₯ 1 π π₯ 2 , π π π₯ 1 < π₯ 2 , πππ‘Γ£π, 3 π₯ 1 β4<3 π₯ 2 β4 π π π₯+β βπ π₯ = 3 π₯+β β4 β 3π₯β4 β3π₯+3ββ4β3π₯+4=3β πΏπππ, π π₯+β βπ π₯ =3β. π΄ ππ₯ππππ π Γ£π ππ πΓ£π πππππππ ππ π, πππ ππ π β΄ π π₯ =3π₯β4 Γ© π’ππ ππ’πçãπ ππππ
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