Pertanyaan

$begin mathsize 14px style stack l i m with t rightwards arrow 2 below fraction numerator 4 t to the power of 4 plus 4 t minus 72 over denominator left parenthesis t minus 2 right parenthesis left parenthesis t squared plus 3 t plus 2 right parenthesis end fraction equals.... end style$

Pembahasan Soal:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell stack l i m with t rightwards arrow 2 below fraction numerator 4 t to the power of 4 plus 4 t minus 72 over denominator left parenthesis t minus 2 right parenthesis left parenthesis t squared plus 3 t plus 2 right parenthesis end fraction end cell equals cell stack l i m with t rightwards arrow 2 below fraction numerator left parenthesis t minus 2 right parenthesis left parenthesis 4 t cubed plus 8 t squared plus 16 t plus 32 right parenthesis over denominator left parenthesis t minus 2 right parenthesis left parenthesis t squared plus 3 t plus 2 right parenthesis end fraction end cell row blank equals cell stack l i m with t rightwards arrow 2 below fraction numerator left parenthesis 4 t cubed plus 8 t squared plus 16 t plus 32 right parenthesis over denominator left parenthesis t squared plus 3 t plus 2 right parenthesis end fraction end cell row blank equals cell fraction numerator 4 left parenthesis 2 right parenthesis cubed plus 8 left parenthesis 2 right parenthesis squared plus 16 left parenthesis 2 right parenthesis plus 32 over denominator 2 squared plus 3 left parenthesis 2 right parenthesis plus 2 end fraction end cell row blank equals cell fraction numerator 32 plus 32 plus 32 plus 32 over denominator 4 plus 6 plus 2 end fraction end cell row blank equals cell 128 over 12 end cell row blank equals cell 32 over 3 end cell end table end style$

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Rochmat

Terakhir diupdate 30 Agustus 2021

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Asimtot datar dari fungsi adalah y= ......

Pembahasan Soal:

Persamaan asimtot datar

Jadi persamaan asimtot datarnya y=2

Jawaban C

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Asimtot tegak dari fungsi adalah ....

Pembahasan Soal:

Perhatikan bahwa bentuk $undefined$ memiliki penyebut = 0 jika x = 2.

Perhatikan bahwa

$begin mathsize 14px style limit as x rightwards arrow 2 to the power of plus of invisible function application fraction numerator 4 x squared minus 3 x plus 7 over denominator x minus 2 end fraction equals limit as x rightwards arrow 2 to the power of plus of invisible function application fraction numerator 4 x squared minus 3 x plus 7 over denominator x minus 2 end fraction times limit as x rightwards arrow 2 to the power of plus of invisible function application fraction numerator 4 x squared minus 3 x plus 7 over denominator x minus 2 end fraction equals limit as x rightwards arrow 2 to the power of plus of invisible function application open parentheses 4 x squared minus 3 x plus 7 close parentheses times limit as x rightwards arrow 2 to the power of plus of invisible function application fraction numerator 1 over denominator x minus 2 end fraction equals open parentheses 4 open parentheses 2 close parentheses squared minus 3 open parentheses 2 close parentheses plus 7 close parentheses times infinity equals open parentheses 16 minus 6 plus 7 close parentheses times infinity equals 17 times infinity equals infinity end style$

Sehingga x = 2 adalah asimtot tegak dari fungsi $begin mathsize 14px style y equals fraction numerator 4 x squared minus 3 x plus 7 over denominator x minus 2 end fraction end style$ .

Roboguru

Berapakah hasil dari ?

Pembahasan Soal:

Pada limit tersebut, saat dilakukan substitusi didapatkan penyebut bernilai nol, sehingga perlu dilakukan pemfaktoran pembilang dan penyebut sebagai berikut,

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x squared minus 4 over denominator x squared minus 3 x plus 2 end fraction end cell equals cell fraction numerator left parenthesis x minus 2 right parenthesis left parenthesis x plus 2 right parenthesis over denominator left parenthesis x minus 2 right parenthesis left parenthesis x minus 1 right parenthesis end fraction end cell row blank equals cell fraction numerator x plus 2 over denominator x minus 1 end fraction end cell end table end style$

sehingga

$stack l i m with x rightwards arrow 2 below fraction numerator x squared minus 4 over denominator x squared minus 3 x plus 2 end fraction equals stack l i m with x rightwards arrow 2 below fraction numerator x plus 2 over denominator x minus 1 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 2 plus 2 over denominator 2 minus 1 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space equals 4 over 1 equals 4$

Jadi, hasil dari $stack l i m with x rightwards arrow 2 below fraction numerator x squared minus 4 over denominator x squared minus 3 x plus 2 end fraction$ adalah 4.

Roboguru