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Asimtot datar dari fungsi adalah ....

Asimtot datar dari fungsi begin mathsize 14px style f open parentheses x close parentheses equals open parentheses x minus 7 close parentheses cubed over open parentheses 4 minus square root of x close parentheses to the power of 6 end style adalah ....
 

  1. x = - 1

  2. x = 1

  3. y = - 1

  4. y = 0

     

  5. y = 1

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H. Nufus

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

jawaban yang tepat adalah E.

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Pembahasan

Perhatikan bahwa Karena maka asimtot datar dari fungsi adalah y = 1. Jadi, jawaban yang tepat adalah E.

Perhatikan bahwa

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 open parentheses x minus 7 close parentheses cubed over open parentheses 4 minus square root of x close parentheses to the power of 6 end cell equals cell limit as x rightwards arrow straight infinity of invisible function application open parentheses open parentheses x minus 7 close parentheses cubed over open parentheses 4 minus square root of x close parentheses to the power of 6 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 open parentheses x minus 7 close parentheses cubed over open parentheses 4 minus square root of x close parentheses to the power of 6 times open parentheses 1 over x close parentheses cubed over open parentheses fraction numerator 1 over denominator square root of x end fraction close parentheses to the power of 6 close parentheses end cell row blank equals cell limit as x rightwards arrow straight infinity of invisible function application open parentheses open parentheses x minus 7 close parentheses times 1 over x close parentheses cubed over open parentheses open parentheses 4 minus square root of x close parentheses times fraction numerator 1 over denominator square root of x end fraction close parentheses to the power of 6 end cell row blank equals cell limit as x rightwards arrow straight infinity of invisible function application open parentheses 1 minus 7 over x close parentheses cubed over open parentheses fraction numerator 4 over denominator square root of x end fraction minus 1 close parentheses to the power of 6 end cell row blank equals cell open parentheses 1 minus 0 close parentheses to the power of 4 over open parentheses 0 minus 1 close parentheses to the power of 6 end cell row blank equals cell 1 to the power of 4 over open parentheses negative 1 close parentheses to the power of 6 end cell row blank equals cell 1 over 1 end cell row blank equals 1 end table end style            

Karena Error converting from MathML to accessible text. maka asimtot datar dari fungsi begin mathsize 14px style f open parentheses x close parentheses equals open parentheses x minus 7 close parentheses cubed over open parentheses 4 minus square root of x close parentheses to the power of 6 end style adalah y = 1.

Jadi, jawaban yang tepat adalah E.

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