Roboguru

Daerah asal fungsi  adalah... .

Pertanyaan

Daerah asal fungsi f open parentheses x close parentheses equals fraction numerator 2 over denominator square root of x squared plus 3 x minus 10 end root end fraction adalah... .

  1. open curly brackets right enclose x minus 2 less than x less than 5 close curly brackets 

  2. open curly brackets right enclose x minus 5 less than x less than 2 close curly brackets 

  3. open curly brackets right enclose x x less than negative 2 space atau space x greater than 5 close curly brackets 

  4. open curly brackets right enclose x x less than 2 space atau space x greater than 5 close curly brackets 

  5. open curly brackets right enclose x x less than negative 5 space atau space x greater than 2 close curly brackets 

Pembahasan Soal:

Diketahui f open parentheses x close parentheses equals fraction numerator 2 over denominator square root of x squared plus 3 x minus 10 end root end fraction. Daerah asal fungsi f open parentheses x close parentheses adalah himpunan x sehingga fungsi f open parentheses x close parentheses terdefinisi. Fungsi f open parentheses x close parentheses equals fraction numerator 2 over denominator square root of x squared plus 3 x minus 10 end root end fraction terdefinisi saat square root of x squared plus 3 x minus 10 end root not equal to 0. Kemudian, square root of x squared plus 3 x minus 10 end root terdefinisi saat x squared plus 3 x minus 10 greater or equal than 0.  Berdasarkan irisan keduanya maka f open parentheses x close parentheses terdefinisi saat x squared plus 3 x minus 10 greater than 0

Nilai x yang memenuhi x squared plus 3 x minus 10 greater than 0 sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus 3 x minus 10 end cell greater than 0 row cell open parentheses x plus 5 close parentheses open parentheses x minus 2 close parentheses end cell greater than 0 end table

open parentheses x plus 5 close parentheses open parentheses x minus 2 close parentheses greater than 0 saat x plus 5 greater than 0 dan x minus 2 greater than 0 atau x plus 5 less than 0 dan x minus 2 less than 0. Nilai x yang memenuhi x plus 5 greater than 0 dan x minus 2 greater than 0 yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 5 end cell greater than 0 row x greater than 5 end table

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 2 end cell greater than 0 row x greater than 2 end table

x greater than 5 dan x greater than 2 maka nilai x yang memenuhi keduanya yaitu x greater than 2. Garis bilangannya sebagai berikut.

Nilai x yang memenuhi  x plus 5 less than 0 dan x minus 2 less than 0 yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 5 end cell less than 0 row x less than cell negative 5 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 2 end cell less than 0 row x less than 2 end table

x less than negative 5 dan x less than 2 maka nilai x yang memenuhi keduanya yaitu x less than negative 5. Garis bilangannya sebagai berikut.

Berdasarkan uraian di atas, fungsi f open parentheses x close parentheses terdefinisi saat x less than negative 5 atau x greater than 2 sehingga D subscript f equals open curly brackets right enclose x x less than negative 5 space atau space x greater than 2 comma space x element of R close curly brackets.

Jadi, jawaban yang benar adalah E.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 04 Mei 2021

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved