Pertanyaan

Asimtot datar dari fungsi adalah ....

Asimtot datar dari fungsi $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator x squared over denominator 3 x minus 2 end fraction minus fraction numerator x squared over denominator 3 x plus 2 end fraction end style$ adalah ....

tidak dapat ditentukan

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

Klaim

H. Nufus

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

jawabannya adalah A.

Pembahasan

Perhatikan bahwa Karena maka asimtot datar dari fungsi adalah . Jadi, jawabannya adalah A.

Perhatikan bahwa

$begin mathsize 14px style limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator x squared over denominator 3 x minus 2 end fraction minus fraction numerator x squared over denominator 3 x plus 2 end fraction close parentheses equals limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator x squared open parentheses 3 x plus 2 close parentheses over denominator open parentheses 3 x minus 2 close parentheses open parentheses 3 x plus 2 close parentheses end fraction minus fraction numerator x squared open parentheses 3 x minus 2 close parentheses over denominator open parentheses 3 x plus 2 close parentheses open parentheses 3 x minus 2 close parentheses end fraction close parentheses equals limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator 3 x cubed plus 2 x squared over denominator 9 x squared minus 4 end fraction minus fraction numerator 3 x cubed minus 2 x squared over denominator 9 x squared minus 4 end fraction close parentheses equals limit as x rightwards arrow straight infinity of invisible function application fraction numerator open parentheses 3 x cubed plus 2 x squared close parentheses minus open parentheses 3 x cubed minus 2 x squared close parentheses over denominator 9 x squared minus 4 end fraction equals limit as x rightwards arrow straight infinity of invisible function application fraction numerator 4 x squared over denominator 9 x squared minus 4 end fraction equals limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator 4 x squared over denominator 9 x squared minus 4 end fraction times fraction numerator 1 over x squared over denominator 1 over x squared end fraction close parentheses equals limit as x rightwards arrow straight infinity of invisible function application fraction numerator 4 over denominator 9 minus 4 over x squared end fraction equals fraction numerator 4 over denominator 9 minus 0 end fraction equals 4 over 9 end style$

Karena maka asimtot datar dari fungsi $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator x squared over denominator 3 x minus 2 end fraction minus fraction numerator x squared over denominator 3 x plus 2 end fraction end style$ adalah .

Jadi, jawabannya adalah A.