Pertanyaan

Misalkan n adalah bilangan bulat sehingga memiliki minimal satu asimtot tegak. Nilai n terkecil yang mungkin adalah ....

Misalkan n adalah bilangan bulat sehingga $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator open parentheses x squared minus 4 close parentheses to the power of 5 open parentheses x squared minus 6 x plus 8 close parentheses squared over denominator open parentheses x minus 2 close parentheses to the power of n end fraction end style$ memiliki minimal satu asimtot tegak. Nilai n terkecil yang mungkin adalah ....

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Klaim

M. Mariyam

Master Teacher

Mahasiswa/Alumni Institut Pertanian Bogor

Jawaban terverifikasi

Pembahasan

Perhatikan bahwa Perhatikan bahwa fungsi f (x) memiliki penyebut = 0 jika x = 2 . Perhatikan bahwa Jika n < 7 , maka 7 - n > 0 , sehingga Maka untuk n < 7 , f( x) tidak memiliki asimtot tegak. Selanjutnya jika n = 7 , maka Maka untuk n = 7 , f( x) tidak memiliki asimtot tegak. Kemudian jika n > 7 , maka n - 7 > 0 , sehingga Maka untuk n > 7 , f( x) memiliki asimtot tegak yaitu x = 2 . Karena n adalah bilangan bulat, maka n terkecil yang mungkin sehingga f (x) memiliki asimtot tegak adalah n = 8 .

Perhatikan bahwa

$begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator open parentheses x squared minus 4 close parentheses to the power of 5 open parentheses x squared minus 6 x plus 8 close parentheses squared over denominator open parentheses x minus 2 close parentheses to the power of n end fraction f open parentheses x close parentheses equals fraction numerator open parentheses open parentheses x minus 2 close parentheses open parentheses x plus 2 close parentheses close parentheses to the power of 5 open parentheses open parentheses x minus 2 close parentheses open parentheses x minus 4 close parentheses close parentheses squared over denominator open parentheses x minus 2 close parentheses to the power of n end fraction f open parentheses x close parentheses equals fraction numerator open parentheses open parentheses x minus 2 close parentheses open parentheses x plus 2 close parentheses close parentheses to the power of 5 open parentheses open parentheses x minus 2 close parentheses open parentheses x minus 4 close parentheses close parentheses squared over denominator open parentheses x minus 2 close parentheses to the power of n end fraction f open parentheses x close parentheses equals fraction numerator open parentheses x minus 2 close parentheses to the power of 5 open parentheses x plus 2 close parentheses to the power of 5 open parentheses x minus 2 close parentheses squared open parentheses x minus 4 close parentheses squared over denominator open parentheses x minus 2 close parentheses to the power of n end fraction f open parentheses x close parentheses equals fraction numerator open parentheses x minus 2 close parentheses to the power of 7 open parentheses x plus 2 close parentheses to the power of 5 open parentheses x minus 4 close parentheses squared over denominator open parentheses x minus 2 close parentheses to the power of n end fraction end style$

Perhatikan bahwa fungsi f(x) 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 open parentheses x minus 2 close parentheses to the power of 7 open parentheses x plus 2 close parentheses to the power of 5 open parentheses x minus 4 close parentheses squared over denominator open parentheses x minus 2 close parentheses to the power of n end fraction equals limit as x rightwards arrow 2 to the power of plus of invisible function application open parentheses x minus 2 close parentheses to the power of 7 over open parentheses x minus 2 close parentheses to the power of n times limit as x rightwards arrow 2 to the power of plus of invisible function application open parentheses open parentheses x plus 2 close parentheses to the power of 5 open parentheses x minus 4 close parentheses squared close parentheses equals limit as x rightwards arrow 2 to the power of plus of invisible function application open parentheses x minus 2 close parentheses to the power of 7 over open parentheses x minus 2 close parentheses to the power of n times open parentheses 2 plus 2 close parentheses to the power of 5 times open parentheses 2 minus 4 close parentheses squared equals limit as x rightwards arrow 2 to the power of plus of invisible function application open parentheses x minus 2 close parentheses to the power of 7 over open parentheses x minus 2 close parentheses to the power of n times 4 to the power of 5 times open parentheses negative 2 close parentheses squared equals limit as x rightwards arrow 2 to the power of plus of invisible function application open parentheses x minus 2 close parentheses to the power of 7 over open parentheses x minus 2 close parentheses to the power of n times 4 to the power of 5 times 4 equals limit as x rightwards arrow 2 to the power of plus of invisible function application open parentheses x minus 2 close parentheses to the power of 7 over open parentheses x minus 2 close parentheses to the power of n times 4 to the power of 6 end style$

Jika n < 7, maka 7 - n > 0, sehingga