Iklan

Pertanyaan

Jika f ( x ) = x 2 − 3 x + 1 , g ( x ) = x + 4 , dan h ( x ) = 2 x − 3 . Nilai ( f ∘ g ∘ h ) ( x ) adalah ....

Jika , dan . Nilai  adalah ....

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

00

:

05

:

08

:

29

Klaim

Iklan

F. Kurnia

Master Teacher

Mahasiswa/Alumni Universitas Jember

Jawaban terverifikasi

Jawaban

hasil dari fungsi tersebut adalah

hasil dari left parenthesis f ring operator g ring operator h right parenthesis left parenthesis x right parenthesis fungsi tersebut adalah Error converting from MathML to accessible text.

Pembahasan

Diketahui , , dan . Ingat bahwa Oleh karena itu, hasil dari fungsi tersebut adalah

Diketahui f left parenthesis x right parenthesis equals x squared minus 3 x plus 1g left parenthesis x right parenthesis equals x plus 4, dan h left parenthesis x right parenthesis equals 2 x minus 3. Ingat bahwa 

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis f ring operator g ring operator h right parenthesis left parenthesis x right parenthesis end cell equals cell f ring operator left parenthesis left parenthesis g ring operator h right parenthesis left parenthesis x right parenthesis right parenthesis end cell row blank equals cell f ring operator left parenthesis g left parenthesis h left parenthesis x right parenthesis right parenthesis right parenthesis end cell row blank equals cell f left parenthesis g left parenthesis h left parenthesis x right parenthesis right parenthesis right parenthesis end cell end table

rightwards arrow left parenthesis g ring operator h right parenthesis left parenthesis x right parenthesis

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis g ring operator h right parenthesis left parenthesis x right parenthesis end cell equals cell g left parenthesis h left parenthesis x right parenthesis right parenthesis end cell row blank equals cell left parenthesis h left parenthesis x right parenthesis right parenthesis plus 4 end cell row blank equals cell left parenthesis 2 x minus 3 right parenthesis plus 4 end cell row blank equals cell 2 x plus 1 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row blank rightwards arrow cell left parenthesis f ring operator g ring operator h right parenthesis left parenthesis x right parenthesis end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis f ring operator g ring operator h right parenthesis left parenthesis x right parenthesis end cell equals cell f ring operator left parenthesis left parenthesis g ring operator h right parenthesis left parenthesis x right parenthesis right parenthesis end cell row blank equals cell f left parenthesis g left parenthesis h left parenthesis x right parenthesis right parenthesis right parenthesis end cell row blank equals cell left parenthesis g left parenthesis h left parenthesis x right parenthesis right parenthesis right parenthesis squared minus 3 left parenthesis g left parenthesis h left parenthesis x right parenthesis right parenthesis right parenthesis plus 1 end cell row blank equals cell left parenthesis 2 x plus 1 right parenthesis squared minus 3 left parenthesis 2 x plus 1 right parenthesis plus 1 end cell row blank equals cell left parenthesis 4 x squared plus 4 x plus 1 right parenthesis minus 6 x minus 3 plus 1 end cell row blank equals cell 4 x squared minus 2 x minus 1 end cell end table

Oleh karena itu, hasil dari left parenthesis f ring operator g ring operator h right parenthesis left parenthesis x right parenthesis fungsi tersebut adalah Error converting from MathML to accessible text.

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

3

Iklan

Pertanyaan serupa

Jika f ( x ) = x , g ( x ) = x + 1 dan h ( x ) = x + 2 maka ( f ∘ g ∘ h ) ( x ) = . . .

2

4.7

Jawaban terverifikasi

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia