Pertanyaan

Diketahui kurva f ( x ) = cos 2 x − 2 cos x − 1 melalui titik berabsis 2 π . Persamaan garis singgung kurva pada titik tersebut adalah...

Diketahui kurva $f (x) = cos^{2} x - 2 cos x - 1$ melalui titik berabsis $\frac{π}{2}$ . Persamaan garis singgung kurva pada titik tersebut adalah...

G. Albiah

Master Teacher

Mahasiswa/Alumni Universitas Galuh Ciamis

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah C.

Pembahasan

Tentukan titik singgung, absis . Ingatlah! Maka, Di dapatkan titik Tentukan turunan pertama dari . Ingatlah! Maka, Kemudian cari turunan Maka, Di dapatkan, Tentukan gradien garis singgungmelalui titik berabsis Ingatlah! Maka, Di dapatkan gradien garis singgung dan titik . Persamaan garis singgung, Jadi, persamaan garis singgung kurva pada titik tersebut adalah . Oleh karena itu, jawaban yang benar adalah C.

Tentukan titik singgung, absis straight pi over 2 . Ingatlah!

cos open parentheses pi over 2 close parentheses equals 0

Maka,

Di dapatkan titik open parentheses x subscript 1 comma y subscript 1 close parentheses equals open parentheses straight pi over 2 comma negative 1 close parentheses

Tentukan turunan pertama dari f left parenthesis x right parenthesis equals cos squared x minus 2 cos x minus 1 . Ingatlah!

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator d f left parenthesis u right parenthesis over denominator d x end fraction end cell equals cell fraction numerator d f over denominator d u end fraction times fraction numerator d u over denominator d x end fraction end cell row f equals cell u squared end cell row u equals cell cos left parenthesis negative 2 cos left parenthesis x right parenthesis minus 1 right parenthesis end cell end table$

Maka,

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator d over denominator d x end fraction open parentheses cos squared open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses close parentheses end cell row blank equals cell fraction numerator d over denominator d u end fraction open parentheses u squared close parentheses fraction numerator d over denominator d x end fraction open parentheses cos open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses close parentheses end cell row blank equals cell 2 u to the power of 2 minus 1 end exponent fraction numerator d over denominator d x end fraction open parentheses cos open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses close parentheses end cell row blank equals cell 2 u fraction numerator d over denominator d x end fraction open parentheses cos open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses close parentheses end cell row blank equals cell 2 cos open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses fraction numerator d over denominator d x end fraction open parentheses cos open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses close parentheses end cell end table$

Kemudian cari turunan $fraction numerator d over denominator d x end fraction open parentheses cos open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses close parentheses$

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator d f left parenthesis u right parenthesis over denominator d x end fraction end cell equals cell fraction numerator d f over denominator d u end fraction times fraction numerator d u over denominator d x end fraction end cell row f equals cell cos space u end cell row u equals cell negative 2 cos left parenthesis x right parenthesis minus 1 end cell end table$

Maka,

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator d over denominator d x end fraction open parentheses cos open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses close parentheses end cell row blank equals cell negative sin open parentheses u close parentheses fraction numerator d over denominator d x end fraction open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses end cell row blank equals cell negative sin open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses times fraction numerator d over denominator d x end fraction open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses end cell row blank equals cell negative sin open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses times negative fraction numerator d over denominator d x end fraction open parentheses 2 cos open parentheses x close parentheses close parentheses minus fraction numerator d over denominator d x end fraction open parentheses 1 close parentheses end cell row blank equals cell negative sin open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses times negative 2 fraction numerator d over denominator d x end fraction open parentheses cos open parentheses x close parentheses close parentheses end cell row blank equals cell negative sin open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses times negative 2 open parentheses negative sin open parentheses x close parentheses close parentheses end cell row blank equals cell negative sin open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses times 2 sin open parentheses x close parentheses end cell end table$

Di dapatkan,

$fraction numerator d over denominator d x end fraction open parentheses cos squared open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses close parentheses equals 2 cos open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses open parentheses negative sin open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses times thin space 2 sin open parentheses x close parentheses close parentheses equals negative 4 cos open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses sin open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses sin open parentheses x close parentheses$

Tentukan gradien garis singgung melalui titik berabsis straight pi over 2

Ingatlah!

$table attributes columnalign right center left columnspacing 0px end attributes row cell sin open parentheses pi over 2 close parentheses end cell equals 1 row cell cos open parentheses pi over 2 close parentheses end cell equals 0 row cell cos open parentheses x close parentheses sin open parentheses x close parentheses end cell equals cell fraction numerator sin open parentheses 2 x close parentheses over denominator 2 end fraction end cell end table$

Maka,

$table attributes columnalign right center left columnspacing 0px end attributes row m equals cell fraction numerator d y over denominator d x end fraction end cell row m equals cell negative 4 cos open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses sin open parentheses negative 2 cos open parentheses x close parentheses minus 1 close parentheses sin open parentheses x close parentheses end cell row blank equals cell negative 4 cos open parentheses negative 2 cos open parentheses straight pi over 2 close parentheses minus 1 close parentheses sin open parentheses negative 2 cos open parentheses straight pi over 2 close parentheses minus 1 close parentheses sin open parentheses straight pi over 2 close parentheses end cell row blank equals cell negative 4 cos open parentheses 2 cos open parentheses pi over 2 close parentheses plus 1 close parentheses sin open parentheses negative 2 cos open parentheses pi over 2 close parentheses minus 1 close parentheses sin open parentheses pi over 2 close parentheses end cell row blank equals cell negative 4 cos open parentheses 2 cos open parentheses pi over 2 close parentheses plus 1 close parentheses open parentheses negative sin open parentheses 2 cos open parentheses pi over 2 close parentheses plus 1 close parentheses close parentheses sin open parentheses pi over 2 close parentheses end cell row blank equals cell 4 sin open parentheses pi over 2 close parentheses cos open parentheses 2 cos open parentheses pi over 2 close parentheses plus 1 close parentheses sin open parentheses 2 cos open parentheses pi over 2 close parentheses plus 1 close parentheses end cell row blank equals cell 4 times thin space 1 times cos open parentheses 2 times thin space 0 plus 1 close parentheses sin open parentheses 2 times thin space 0 plus 1 close parentheses end cell row blank equals cell 4 cos open parentheses 1 close parentheses sin open parentheses 1 close parentheses end cell row blank equals cell 4 times fraction numerator sin open parentheses 2 times thin space 1 close parentheses over denominator 2 end fraction end cell row blank equals cell 4 times fraction numerator sin open parentheses 2 close parentheses over denominator 2 end fraction end cell row blank equals cell 2 sin open parentheses 2 close parentheses end cell row blank equals cell 1 comma 8 end cell row blank almost equal to 2 end table$

Di dapatkan gradien garis singgung m equals 2 dan titik open parentheses x subscript 1 comma y subscript 1 close parentheses equals open parentheses straight pi over 2 comma negative 1 close parentheses . Persamaan garis singgung,

$table attributes columnalign right center left columnspacing 0px end attributes row cell y minus y subscript 1 end cell equals cell m open parentheses x minus x subscript 1 close parentheses end cell row cell y minus left parenthesis negative 1 right parenthesis end cell equals cell 2 open parentheses x minus straight pi over 2 close parentheses end cell row cell y plus 1 end cell equals cell 2 open parentheses x minus straight pi over 2 close parentheses end cell row y equals cell 2 open parentheses x minus straight pi over 2 close parentheses minus 1 end cell row y equals cell 2 x minus fraction numerator 2 straight pi over denominator 2 end fraction minus 1 end cell row y equals cell 2 x minus straight pi minus 1 end cell end table$

Jadi, persamaan garis singgung kurva pada titik tersebut adalah y equals 2 x minus straight pi minus 1 .

Oleh karena itu, jawaban yang benar adalah C.