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Bentuk sederhana dari  untuk nilai  adalah ....

Pertanyaan

Bentuk sederhana dari open vertical bar 3 x minus 6 close vertical bar minus open vertical bar x minus 4 close vertical bar cross times open vertical bar x plus 1 close vertical bar untuk nilai 2 less than x less than 4 adalah ....

  1. x squared plus 2

  2. x squared minus 10

  3. x squared minus 6 x plus 10

  4. x squared minus 6 x plus 2

  5. x squared minus 6 x minus 2

Pembahasan Soal:

Ingat bahwa definisi nilai mutlak yaitu

open vertical bar f left parenthesis x right parenthesis close vertical bar open curly brackets table attributes columnalign left end attributes row cell f left parenthesis x right parenthesis comma space space space space space untuk space f left parenthesis x right parenthesis greater or equal than 0 end cell row cell negative f left parenthesis x right parenthesis comma space space space space space untuk space f left parenthesis x right parenthesis less than 0 end cell end table close

Berdasarkan definisi tersebut maka 

open vertical bar 3 x minus 6 close vertical bar open curly brackets table attributes columnalign left end attributes row cell 3 x minus 6 comma space space space space space untuk space 3 x minus 6 greater or equal than 0 end cell row cell space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space x greater or equal than 2 end cell row cell negative left parenthesis 3 x minus 6 right parenthesis comma space space space space space untuk space 3 x minus 6 less than 0 end cell row cell space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space x less than 2 end cell end table close

open vertical bar x minus 4 close vertical bar open curly brackets table attributes columnalign left end attributes row cell x minus 4 comma space space space space space untuk space x minus 4 greater or equal than 0 end cell row cell space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space x greater or equal than 4 end cell row cell negative left parenthesis x minus 4 right parenthesis comma space space space space space untuk space x minus 4 less than 0 end cell row cell space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space x less than 4 end cell end table close

open vertical bar x plus 1 close vertical bar open curly brackets table attributes columnalign left end attributes row cell x plus 1 comma space space space space space untuk space x plus 1 greater or equal than 0 end cell row cell space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space x greater or equal than negative 1 end cell row cell negative left parenthesis x plus 1 right parenthesis comma space space space space space untuk space x plus 1 less than 0 end cell row cell space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space x less than negative 1 end cell end table close

 

Sehingga bentuk sederhana dari open vertical bar 3 x minus 6 close vertical bar minus open vertical bar x minus 4 close vertical bar cross times open vertical bar x plus 1 close vertical bar untuk nilai 2 less than x less than 4 adalah

rightwards arrow open vertical bar 3 x minus 6 close vertical bar minus open vertical bar x minus 4 close vertical bar cross times open vertical bar x plus 1 close vertical bar equals 3 x plus 6 minus left parenthesis negative left parenthesis x minus 4 right parenthesis right parenthesis cross times left parenthesis x plus 1 right parenthesis equals 3 x plus 6 plus open parentheses left parenthesis x minus 4 right parenthesis cross times left parenthesis x plus 1 right parenthesis close parentheses equals 3 x plus 6 plus left parenthesis x squared minus 4 x plus x minus 4 right parenthesis equals 3 x plus 6 plus x squared minus 3 x minus 4 equals x squared plus 2

Oleh karena itu, jawaban yang benar adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Kurnia

Mahasiswa/Alumni Universitas Jember

Terakhir diupdate 12 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Nilai-nilai x yang memenuhi  adalah ....

Pembahasan Soal:

Ingat kembali:

Nilai mutlak didefinisikan sebagai berikut 

open vertical bar x close vertical bar equals open curly brackets table attributes columnalign right center columnspacing 1.4ex end attributes row cell x comma end cell cell text untuk end text space x greater or equal than 0 end cell row cell negative x comma end cell cell text untuk end text space x less than 0 end cell end table close

Oleh karena bentuk mutlak dari pertidaksamaan adalah open vertical bar 1 minus 2 x close vertical bar, maka diperoleh 

open vertical bar 1 minus 2 x close vertical bar equals open curly brackets table attributes columnspacing 1.4ex end attributes row cell 1 minus 2 x comma end cell cell text untuk end text space 1 minus 2 x greater or equal than 0 end cell row blank cell text atau end text space space space space space space space space space space x less or equal than 1 half end cell row blank blank row cell negative open parentheses 1 minus 2 x close parentheses equals negative 1 plus 2 x comma end cell cell text untuk end text space 1 minus 2 x less than 0 end cell row blank cell text atau end text space space space space space space space space space space x greater than 1 half end cell end table close

Untuk x less or equal than 1 half, pertidaksamaan yang diselesaikan adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 2 end cell less or equal than cell 1 minus 2 x end cell row cell x plus 2 x end cell less or equal than cell 1 plus 2 end cell row cell 3 x end cell less or equal than 3 row x less or equal than cell 3 over 3 end cell row x less or equal than 1 end table

sehingga solusi pertidaksamaan untuk x less or equal than 1 half adalah x less or equal than 1 half intersection space x less or equal than 1 equals x less or equal than 1 half.

 Untuk x greater than 1 half, pertidaksamaan yang diselesaikan adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 2 end cell less or equal than cell negative open parentheses 1 minus 2 x close parentheses end cell row cell x minus 2 end cell less or equal than cell negative 1 plus 2 x end cell row cell x minus 2 x end cell less or equal than cell negative 1 plus 2 end cell row cell negative x end cell less or equal than 1 row x greater or equal than cell negative 1 end cell end table

sehingga solusi pertidaksamaan untuk x greater than 1 half adalah x greater than 1 half intersection space x greater or equal than negative 1 equals x greater than 1 half.

Dengan demikian, karena diperoleh solusi pertidaksamaan adalah x less or equal than 1 half dan x greater than 1 half, maka solusi dari x minus 2 less or equal than open vertical bar 1 minus 2 x close vertical baradalah x less or equal than 1 half union x greater than 1 half equals text semua bilangan riil end text.

Oleh karena itu, jawaban yang benar adalah A.space 

1

Roboguru

Petunjuk: Gunakan kalkulator untuk menentukan nilai yang diminta berikut. Jika  tentukanlah

Pembahasan Soal:

Ingat definisi fungsi nilai mutlak f open parentheses x close parentheses equals open vertical bar x close vertical bar yang berarti bahwa fungsi akan selalu bernilai greater or equal than 0 sehingga dapat ditulis:

f open parentheses x close parentheses equals open curly brackets table attributes columnalign left left left end attributes row x untuk cell x greater or equal than 0 end cell row cell negative x end cell untuk cell x less than 0 end cell end table close  

Diketahui f open parentheses x close parentheses equals open vertical bar 0 comma 175 x plus 9 comma 762 close vertical barmaka dapat ditulis:

f open parentheses x close parentheses equals open curly brackets table row cell 0 comma 175 x plus 9 comma 762 end cell untuk cell x greater or equal than negative 55 comma 78 end cell row cell negative open parentheses 0 comma 175 x plus 9 comma 762 close parentheses end cell untuk cell x less than negative 55 comma 78 end cell end table close 

f open parentheses negative 0 comma 123 close parentheses oleh karena x greater or equal than negative 55 comma 78 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell 0 comma 175 x plus 9 comma 762 end cell row cell f open parentheses negative 0 comma 123 close parentheses end cell equals cell 0 comma 175 open parentheses negative 0 comma 123 close parentheses plus 9 comma 762 end cell row blank equals cell negative 0 comma 021525 plus 9 comma 762 end cell row blank equals cell 9 comma 74 end cell end table 

Jadi nilai dari f open parentheses negative 0 comma 123 close parentheses adalah 9 comma 74.

0

Roboguru

Petunjuk: Gunakan kalkulator untuk menentukan nilai yang diminta berikut. Jika  tentukanlah .

Pembahasan Soal:

Ingat definisi fungsi nilai mutlak f open parentheses x close parentheses equals open vertical bar x close vertical bar yang berarti bahwa fungsi akan selalu bernilai greater or equal than 0 sehingga dapat ditulis:

f open parentheses x close parentheses equals open curly brackets table attributes columnalign left left left end attributes row x untuk cell x greater or equal than 0 end cell row cell negative x end cell untuk cell x less than 0 end cell end table close  

Diketahui f open parentheses x close parentheses equals open vertical bar 753 x minus 39.543 close vertical bar maka dapat ditulis:

f open parentheses x close parentheses equals open curly brackets table row cell 753 x minus 39.543 end cell untuk cell x greater or equal than 52 comma 51 end cell row cell negative open parentheses 753 x minus 39.543 close parentheses end cell untuk cell x less than 52 comma 51 end cell end table close  

f open parentheses 0 comma 245 close parentheses oleh karena x less than 0 comma 245 maka gunakan f open parentheses x close parentheses equals negative open parentheses 753 x minus 39.543 close parentheses sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell negative open parentheses 753 x minus 39.543 close parentheses end cell row cell f open parentheses 0 comma 245 close parentheses end cell equals cell negative open parentheses 753 open parentheses 0 comma 245 close parentheses minus 39.543 close parentheses end cell row blank equals cell negative open parentheses 184 comma 485 minus 39.543 close parentheses end cell row blank equals cell negative open parentheses negative 39.358 comma 515 close parentheses end cell row blank equals cell 39.358 comma 515 end cell end table  

Jadi nilai dari f open parentheses 0 comma 245 close parentheses adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 39 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank. end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 358 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank comma end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 515 end table.

0

Roboguru

Petunjuk: Gunakan kalkulator untuk menentukan nilai yang diminta berikut. Jika  tentukanlah nilai  yang memenuhi untuk .

Pembahasan Soal:

Ingat definisi nilai mutlak, misalkan x bilangan real, open vertical bar x close vertical bar dibaca nilai mutlak x dan didefinisikan:

open vertical bar x close vertical bar equals open curly brackets table row x jika cell x greater or equal than 0 end cell row cell negative x end cell jika cell x less than 0 end cell end table close 

atau 

open vertical bar x close vertical bar equals open curly brackets table row cell f open parentheses x close parentheses end cell jika cell f open parentheses x close parentheses greater or equal than 0 end cell row cell negative f open parentheses x close parentheses end cell jika cell f open parentheses x close parentheses less than 0 end cell end table close 

Diketahui f open parentheses x close parentheses equals open vertical bar 375 x minus 519 comma 42 close vertical bar dengan f open parentheses x close parentheses equals 334 comma 5 maka dapat ditulis open vertical bar 375 x minus 519 comma 42 close vertical bar equals 334 comma 5 sehingga:

begin mathsize 12px style open vertical bar 375 x minus 519 comma 42 close vertical bar equals open curly brackets table row cell open parentheses 375 x minus 519 comma 42 close parentheses end cell jika cell 375 x minus 519 comma 42 greater or equal than 0 end cell row cell negative open parentheses 375 x minus 519 comma 42 close parentheses end cell jika cell 375 x minus 519 comma 42 less than 0 end cell end table close end style  

Jika 375 x minus 519 comma 42 greater or equal than 0 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 375 x minus 519 comma 42 close vertical bar end cell equals cell 334 comma 5 end cell row cell 375 x minus 519 comma 42 end cell equals cell 334 comma 5 end cell row cell 375 x end cell equals cell 334 comma 5 plus 519 comma 42 end cell row cell 375 x end cell equals cell 853 comma 92 end cell row x equals cell 2 comma 28 end cell end table 

jika 375 x minus 519 comma 42 less than 0 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 375 x minus 519 comma 42 close vertical bar end cell equals cell 334 comma 5 end cell row cell negative open parentheses 375 x minus 519 comma 42 close parentheses end cell equals cell 334 comma 5 end cell row cell negative 375 x plus 519 comma 42 end cell equals cell 334 comma 5 end cell row cell 375 x end cell equals cell 334 comma 5 minus 519 comma 42 end cell row cell 375 x end cell equals cell negative 184 comma 92 end cell row x equals cell negative 0 comma 49 end cell end table 

 Dengan demikian nilai x yang memenuhi adalah x equals 2 comma 28 dan x equals negative 0 comma 49.

0

Roboguru

Tentukan nilai  yang memenuhi

Pembahasan Soal:

Ingat definisi nilai mutlak, misalkan x bilangan real, open vertical bar x close vertical bar dibaca nilai mutlak x, dan didefinisikan:

open vertical bar x close vertical bar equals open curly brackets table row x jika cell x greater or equal than 0 end cell row cell negative x end cell jika cell x less than 0 end cell end table close 

Diketahui open vertical bar 0 comma 175 x plus 9 comma 762 close vertical bar equals 123 maka dapat ditulis:

begin mathsize 12px style open vertical bar 0 comma 175 x plus 9 comma 762 close vertical bar equals open curly brackets table row cell 0 comma 175 x plus 9 comma 762 end cell jika cell 0 comma 175 x plus 9 comma 762 greater or equal than 0 end cell row cell negative open parentheses 0 comma 175 x plus 9 comma 762 close parentheses end cell jika cell 0 comma 175 x plus 9 comma 762 less than 0 end cell end table close end style 

Untuk 0 comma 175 x plus 9 comma 762 greater or equal than 0 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 0 comma 175 x plus 9 comma 762 close vertical bar end cell equals 123 row cell 0 comma 175 x plus 9 comma 762 end cell equals 123 row cell 0 comma 175 x end cell equals cell 123 minus 9 comma 762 end cell row cell 0 comma 175 x end cell equals cell 113 comma 238 end cell row x equals cell fraction numerator 113 comma 238 over denominator 0 comma 175 end fraction end cell row x equals cell 647 comma 07 end cell end table  

Untuk 0 comma 175 x plus 9 comma 762 less than 0 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 0 comma 175 x plus 9 comma 762 close vertical bar end cell equals 123 row cell negative open parentheses 0 comma 175 x plus 9 comma 762 close parentheses end cell equals 123 row cell negative 0 comma 175 x minus 9 comma 762 end cell equals 123 row cell negative 0 comma 175 x end cell equals cell 123 plus 9 comma 762 end cell row cell negative 0 comma 175 x end cell equals cell 132 comma 762 end cell row x equals cell fraction numerator 132 comma 762 over denominator negative 0 comma 175 end fraction end cell row x equals cell negative 758 comma 64 end cell end table 

Dengan demikian nilai x yang memenuhi open vertical bar 0 comma 175 x plus 9 comma 762 close vertical bar equals 123 adalah x equals 647 comma 07 dan x equals negative 758 comma 64.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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