Pertanyaan

Asimtot datar dari fungsi adalah ....

Asimtot datar dari fungsi $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator open parentheses 3 x plus 7 close parentheses cubed over denominator 5 x cubed minus 3 x end fraction end style$ adalah ....

D. Natalia

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

Ingatbahwa asimtot datar dari adalah . Nilai dari dapat dicari sebagai berikut. Karena ,maka asimtot datar dari fungsi adalah . Jadi, jawaban yang tepat adalah B.

Ingat bahwa asimtot datar dari $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator open parentheses 3 x plus 7 close parentheses cubed over denominator 5 x cubed minus 3 x end fraction end style$ adalah $begin mathsize 14px style straight y equals limit as straight x rightwards arrow infinity of fraction numerator open parentheses 3 straight x plus 7 close parentheses cubed over denominator 5 straight x cubed minus 3 straight x end fraction end style$ .

Nilai dari $begin mathsize 14px style limit as straight x rightwards arrow infinity of fraction numerator open parentheses 3 straight x plus 7 close parentheses cubed over denominator 5 straight x cubed minus 3 straight x end fraction end style$ dapat dicari sebagai berikut.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow straight infinity of invisible function application fraction numerator open parentheses 3 x plus 7 close parentheses cubed over denominator 5 x cubed minus 3 x end fraction end cell equals cell limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator open parentheses 3 x plus 7 close parentheses cubed over denominator 5 x cubed minus 3 x end fraction times fraction numerator 1 over x cubed over denominator 1 over x cubed end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator open parentheses 3 x plus 7 close parentheses cubed over denominator 5 x cubed minus 3 x end fraction times fraction numerator open parentheses 1 over x close parentheses cubed over denominator 1 over x cubed end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow straight infinity of invisible function application fraction numerator open parentheses open parentheses 3 x plus 7 close parentheses times 1 over x close parentheses cubed over denominator 5 minus 3 over x squared end fraction end cell row blank equals cell limit as x rightwards arrow straight infinity of invisible function application fraction numerator open parentheses 3 plus 7 over x close parentheses cubed over denominator 5 minus 3 over x squared end fraction end cell row blank equals cell fraction numerator open parentheses 3 plus 0 close parentheses cubed over denominator 5 minus 0 end fraction end cell row blank equals cell 3 cubed over 5 end cell row blank equals cell 27 over 5 end cell end table end style$

Karena $begin mathsize 14px style limit as straight x rightwards arrow straight infinity of invisible function application fraction numerator open parentheses 3 straight x plus 7 close parentheses cubed over denominator 5 straight x cubed minus 3 straight x end fraction equals 27 over 5 end style$ , maka asimtot datar dari fungsi $begin mathsize 14px style straight f open parentheses straight x close parentheses equals fraction numerator open parentheses 3 straight x plus 7 close parentheses cubed over denominator 5 straight x cubed minus 3 straight x end fraction end style$ adalah y equals 27 over 5 .

Jadi, jawaban yang tepat adalah B.