Question

MATEMATICA AJUDA
Sejam A= [1,2] e B=[0,1,2]

Answer 1

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☞ (c) f : x ⇒ x² -3x + 2 em uma função de A em B. ✅

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$\green{\rm\underline{EXPLICAC_{\!\!\!,}\tilde{A}O~PASSO{-}A{-}PASSO~~~}}$✍

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☔ Oi, Thomas. Inicialmente  lembremos que:

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• Função é uma regra que relaciona cada elemento de um conjunto A (chamado Domínio) a um único elemento de outro conjunto B (chamado Imagem).

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☔ Verificaremos, portanto, se os elementos do domínio de cada item possuem uma, e somente uma, imagem.

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Ⓐ$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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$\LARGE\gray{\boxed{\sf\blue{~~f:~\overbrace{\sf x}^{A: Dm}~\rightarrow~\overbrace{\sf 2x}^{B: Im}~~}}}$

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➡ x = 1

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$\Large\blue{\sf 2x = 2 \cdot 1 = 2~\longrightarrow~\{2 \in B\} }$  ✅

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➡ x = 2

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$\Large\blue{\sf 2x = 2 \cdot 2 = 4~~\longrightarrow~\{~4 \notin B\} }$  ❌

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Ⓑ$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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$\LARGE\gray{\boxed{\sf\blue{~~f:~\overbrace{\sf x}^{A: Dm}~\rightarrow~\overbrace{\sf x + 1}^{B: Im}~~}}}$

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➡ x = 1

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$\Large\blue{\sf x + 1 = 1 + 1 = 2~~\longrightarrow~\{~2 \in B\} }$  ✅

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➡ x = 2

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$\Large\blue{\sf x + 1 = 2 + 1 = 3~~\longrightarrow~\{~3 \notin B\} }$  ❌

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Ⓒ$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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$\LARGE\gray{\boxed{\sf\blue{~~f:~\overbrace{\sf x}^{A: Dm}~\rightarrow~\overbrace{\sf x^2 - 3x + 2}^{B: Im}~~}}}$

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➡ x = 1

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$\Large\blue{\sf x^2 - 3x + 2 = 1^2 - 3 \cdot 1 + 2 = 0~~\longrightarrow~\{~1 \in B\} }$  ✅

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➡ x = 2

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$\Large\blue{\sf x^2 - 3x + 2 = 2^2 - 3 \cdot 2 + 2 = 0~~\longrightarrow~\{~0 \in B\} }$  ✅

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Ⓓ$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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$\LARGE\gray{\boxed{\sf\blue{~~f:~\overbrace{\sf x}^{B: Dm}~\rightarrow~\overbrace{\sf x^2 - x}^{A: Im}~~}}}$

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➡ x = 0

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$\Large\blue{\sf x^2 - x = 0^2 - 0 = 0~~\longrightarrow~\{~0 \notin A\} }$  ❌

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➡  x = 1

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$\Large\blue{\sf x^2 - x = 1^2 - 1 = 0~~\longrightarrow~\{~0 \notin A\} }$  ❌

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➡  x = 2

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$\Large\blue{\sf x^2 - x = 2^2 - 2 = 2~~\longrightarrow~\{~2 \in A\} }$  ✅

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Ⓔ$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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$\LARGE\gray{\boxed{\sf\blue{~~f:~\overbrace{\sf x}^{B: Dm}~\rightarrow~\overbrace{\sf x - 1}^{A: Im}~~}}}$

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➡  x = 0

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$\Large\blue{\sf x - 1 = 0 - 1 = -1~~\longrightarrow~\{-1 \notin A\} }$  ❌

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➡  x = 1

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$\Large\blue{\sf x - 1 = 1 - 1 = 0~~\longrightarrow~\{~0 \notin A\} }$  ❌

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➡ x = 2

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$\Large\blue{\sf x - 1 = 2 - 1 = 1~~\longrightarrow~\{~2 \in A\} }$  ✅

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$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$☁

☕ $\bf\Large\blue{Bons\ estudos.}$

($\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios}$) ☄

$\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }}\LaTeX$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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$\gray{"Absque~sudore~et~labore~nullum~opus~perfectum~est."}$