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Daerah asal grafik fungsi y = f ( x ) = x 2 + 1 dengan daerah hasil R f ​ = { x ∣ 5 ≤ y < 10 , y ∈ R } adalah ....

Daerah asal grafik fungsi dengan daerah hasil adalah

  1. begin mathsize 14px style open curly brackets right enclose x space minus 3 less or equal than x less than 3 comma space x element of R close curly brackets end style

  2. begin mathsize 14px style open curly brackets right enclose x space minus 3 less than x less than 3 comma space x element of R close curly brackets end style

  3. begin mathsize 14px style open curly brackets right enclose x space minus 3 less or equal than x less than negative 2 comma space dan space 2 less or equal than x less than 3 comma space x element of R close curly brackets end style

  4. begin mathsize 14px style open curly brackets right enclose x space minus 3 less or equal than x less than negative 2 comma space dan space 2 less than x less or equal than 3 comma space x element of R close curly brackets end style

  5. begin mathsize 14px style open curly brackets right enclose x space minus 3 less than x less or equal than negative 2 comma space dan space 2 less or equal than x less than 3 comma space x element of R close curly brackets end style

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F. Kartikasari

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

jawaban yang tepat adalah E.

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Pembahasan

Sehingga, . Jadi, jawaban yang tepat adalah E.

begin mathsize 14px style 5 less or equal than y less than 10 5 less or equal than x squared plus 1 less than 10 4 less or equal than x squared less than 9 end style

begin mathsize 14px style x squared minus 4 greater or equal than 0 space dan space x squared minus 9 less than 0 end style

Sehingga, begin mathsize 14px style straight D subscript straight f equals open curly brackets right enclose x space minus 3 less than x less or equal than negative 2 space dan space 2 less or equal than x less than 3 comma space x element of straight R close curly brackets end style.

Jadi, jawaban yang tepat adalah E.

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Muhammad Fathir Rahman Pradana

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