Pertanyaan

Tentukan interval x sehingga grafik f ( x ) = − 2 sin ( 2 x − 6 π ) cekung ke bawah untuk 0 ≤ x ≤ π !

Tentukan interval $x$ sehingga grafik $f (x) = - 2 sin (2 x - \frac{π}{6})$ cekung ke bawah untuk $0 \leq x \leq π$ !

D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Jawaban

grafik cekung ke bawah pada interval .

grafik

f left parenthesis x right parenthesis equals negative 2 space cos space open parentheses 2 x minus straight pi over 6 close parentheses

cekung ke bawah pada interval $table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight pi over 3 end cell end table less than x less than table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 5 straight pi over denominator 6 end fraction end cell end table$ .

Pembahasan

Cekung ke atas atau ke bawah suatu tungsi dalam suatu interval tertentu dapat dilihat dari turunan keduanya. Jika , fungsi cekung ke atas. Jika ,fungsi cekung ke bawah. Diketahui , maka sedangkan . Fungsi cekung ke bawah ketika , maka : Nilai perbandingan kosinus bertanda negatif di kuadran II dan IIIsehingga : dan Fungsi memiliki interval , sehingga interval positif fungsi terdapat pada . Sehingga grafik cekung ke bawah pada interval .

Cekung ke atas atau ke bawah suatu tungsi dalam suatu interval tertentu dapat dilihat dari turunan keduanya.

Jika , fungsi cekung ke atas.
Jika , fungsi cekung ke bawah.

Diketahui f left parenthesis x right parenthesis equals negative 2 space cos space open parentheses 2 x minus straight pi over 6 close parentheses , maka f apostrophe left parenthesis x right parenthesis equals 4 space sin space open parentheses 2 x minus straight pi over 6 close parentheses sedangkan f apostrophe apostrophe left parenthesis x right parenthesis equals 8 space cos space open parentheses 2 x minus straight pi over 6 close parentheses .

Fungsi cekung ke bawah ketika f apostrophe apostrophe left parenthesis x right parenthesis less than 0 , maka :

table attributes columnalign right center left columnspacing 0px end attributes row cell f apostrophe apostrophe left parenthesis x right parenthesis end cell less than 0 row cell 8 space cos space open parentheses 2 x minus straight pi over 6 close parentheses end cell less than 0 row cell cos space open parentheses 2 x minus straight pi over 6 close parentheses end cell less than 0 end table

Nilai perbandingan kosinus bertanda negatif di kuadran II dan III sehingga :

$table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses 2 x minus straight pi over 6 close parentheses end cell greater than cell cos space straight pi over 2 end cell row cell 2 x minus straight pi over 6 end cell greater than cell straight pi over 2 end cell row cell 2 x end cell greater than cell straight pi over 2 plus straight pi over 6 end cell row cell 2 x end cell greater than cell fraction numerator 3 straight pi over denominator 6 end fraction plus straight pi over 6 end cell row cell 2 x end cell greater than cell fraction numerator 4 straight pi over denominator 6 end fraction end cell row x greater than cell fraction numerator 2 straight pi over denominator 6 end fraction end cell row x greater than cell straight pi over 3 end cell end table$

dan

$table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses 2 x minus straight pi over 6 close parentheses end cell less than cell cos space fraction numerator 3 straight pi over denominator 2 end fraction end cell row cell 2 x minus straight pi over 6 end cell less than cell fraction numerator 3 straight pi over denominator 2 end fraction end cell row cell 2 x end cell less than cell fraction numerator 3 straight pi over denominator 2 end fraction plus straight pi over 6 end cell row cell 2 x end cell less than cell fraction numerator 9 straight pi over denominator 6 end fraction plus straight pi over 6 end cell row cell 2 x end cell less than cell fraction numerator 10 straight pi over denominator 6 end fraction end cell row x less than cell fraction numerator 5 straight pi over denominator 6 end fraction end cell end table$

Fungsi memiliki interval 0 less or equal than x less or equal than straight pi , sehingga interval positif fungsi terdapat pada $table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight pi over 3 end cell end table less than x less than table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 5 straight pi over denominator 6 end fraction end cell end table$ .

Sehingga grafik f left parenthesis x right parenthesis equals negative 2 space cos space open parentheses 2 x minus straight pi over 6 close parentheses cekung ke bawah pada interval $table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight pi over 3 end cell end table less than x less than table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 5 straight pi over denominator 6 end fraction end cell end table$ .