Daerah hasil fungsi f(x)=−6sin(x−3π​)+1 adalah ....

Pertanyaan

Daerah hasil fungsi f left parenthesis x right parenthesis equals negative 6 sin open parentheses x minus straight pi over 3 close parentheses plus 1 adalah ....

  1. open curly brackets y vertical line minus 7 less or equal than y less or equal than 5 comma space y element of R close curly brackets 

  2. open curly brackets y vertical line minus 6 less or equal than y less or equal than 6 comma space y element of R close curly brackets  

  3. open curly brackets y vertical line 5 less or equal than y less or equal than 7 comma space y element of R close curly brackets 

  4. open curly brackets y vertical line minus 5 less or equal than y less or equal than 5 comma space y element of R close curly brackets 

  5. open curly brackets y vertical line minus 5 less or equal than y less or equal than 7 comma space y element of R close curly brackets 

I. Kumaralalita

Master Teacher

Mahasiswa/Alumni Universitas Gadjah Mada

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

Pembahasan

Diberikan fungsi trigonometri f left parenthesis x right parenthesis equals negative 6 sin open parentheses x minus straight pi over 3 close parentheses plus 1. Nilai maksimum dan minimum fungsi merupakan batas atas dan batas bawah daerah hasil atau range. Nilai maksimum dan minimum dapat dicari dengan turunan fungsi.

f apostrophe left parenthesis x right parenthesis equals negative 6 space cos open parentheses x minus pi over 3 close parentheses.

Selanjutnya tentukan x dari syarat stasioner :

table attributes columnalign right center left columnspacing 0px end attributes row cell f apostrophe left parenthesis x right parenthesis end cell equals 0 row cell negative 6 space cos open parentheses x minus straight pi over 3 close parentheses end cell equals 0 row cell cos open parentheses x minus straight pi over 3 close parentheses end cell equals cell cos space straight pi over 2 end cell end table 

Dalam radian, persamaan trigonometri cosinus dapat diselesaikan seperti berikut :

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space x end cell equals cell cos space alpha end cell row x equals cell plus-or-minus alpha plus k times 2 straight pi end cell end table 

dengan k adalah bilangan bulat. Oleh karena itu, penyelesaian persamaan di atas adalah 

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus straight pi over 3 end cell equals cell plus-or-minus straight pi over 2 plus k times 2 straight pi end cell row straight x equals cell plus-or-minus straight pi over 2 plus straight pi over 3 plus straight k times 2 straight pi end cell end table 

Ditetapkan domain fungsi adalah D equals open curly brackets 0 less or equal than x less or equal than 2 straight pi close curly brackets, maka untuk x equals straight pi over 2 plus straight pi over 3 plus k times 2 straight pi equals fraction numerator 5 straight pi over denominator 6 end fraction plus straight k times 2 straight pi :

k equals 0 rightwards arrow x equals fraction numerator 5 straight pi over denominator 6 end fraction 

Untuk x equals negative straight pi over 2 plus straight pi over 3 plus k times 2 straight pi equals negative straight pi over 6 plus straight k times 2 straight pi :

k equals 1 rightwards arrow x equals 2 straight pi minus straight pi over 6 equals fraction numerator 11 straight pi over denominator 6 end fraction

Didapatkan x subscript 1 equals fraction numerator 5 straight pi over denominator 6 end fraction dan x subscript 2 equals fraction numerator 11 straight pi over denominator 6 end fraction. Substitusikan ke dalam fungsi awal :

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses fraction numerator 5 straight pi over denominator 6 end fraction close parentheses end cell equals cell negative 6 space sin open parentheses fraction numerator 5 straight pi over denominator 6 end fraction minus straight pi over 3 close parentheses plus 1 end cell row blank equals cell negative 6 space sin open parentheses fraction numerator 5 straight pi minus 2 straight pi over denominator 6 end fraction close parentheses plus 1 end cell row blank equals cell negative 6 open parentheses sin space fraction numerator 3 straight pi over denominator 6 end fraction close parentheses plus 1 end cell row blank equals cell negative 6 open parentheses sin space straight pi over 2 close parentheses plus 1 end cell row blank equals cell negative 6 left parenthesis 1 right parenthesis plus 1 end cell row blank equals cell negative 6 plus 1 end cell row blank equals cell negative 5 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses fraction numerator 11 straight pi over denominator 6 end fraction close parentheses end cell equals cell negative 6 space sin open parentheses fraction numerator 11 straight pi over denominator 6 end fraction minus straight pi over 3 close parentheses plus 1 end cell row blank equals cell negative 6 space sin open parentheses fraction numerator 11 straight pi minus 2 straight pi over denominator 6 end fraction close parentheses plus 1 end cell row blank equals cell negative 6 open parentheses sin space fraction numerator 9 straight pi over denominator 6 end fraction close parentheses plus 1 end cell row blank equals cell negative 6 open parentheses sin space fraction numerator 3 straight pi over denominator 2 end fraction close parentheses plus 1 end cell row blank equals cell negative 6 left parenthesis negative 1 right parenthesis plus 1 end cell row blank equals cell 6 plus 1 end cell row blank equals 7 end table 

Didapatkan nilai minimum dan maksimum fungsi adalah negative 5 space dan space 7. Dengan demikian, daerah hasil fungsi f left parenthesis x right parenthesis adalah open curly brackets y vertical line minus 5 less or equal than y less or equal than 7 comma space y element of R close curly brackets.

Jadi, jawaban yang tepat adalah E.

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