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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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M. Mariyam

Master Teacher

Mahasiswa/Alumni Institut Pertanian Bogor

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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

begin mathsize 14px style 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 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 minus n end exponent v 4 to the power of 6 equals open parentheses 2 minus 2 close parentheses to the power of 7 minus n end exponent times 4 to the power of 6 equals 0 to the power of 7 minus n end exponent times 4 to the power of 6 equals 0 end style    

Maka untuk n < 7, f(x) tidak memiliki asimtot tegak.

 

Selanjutnya jika n = 7, maka

begin mathsize 14px style 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 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 7 times 4 to the power of 6 equals limit as x rightwards arrow 2 to the power of plus of invisible function application 4 to the power of 6 equals 4 to the power of 6 end style    

Maka untuk n = 7, f(x) tidak memiliki asimtot tegak.

 

Kemudian jika n > 7, maka n - 7 > 0 , sehingga

begin mathsize 14px style 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 equals limit as x rightwards arrow 2 to the power of plus of invisible function application 1 over open parentheses x minus 2 close parentheses to the power of n minus 7 end exponent times 4 to the power of 6 equals infinity times 4 to the power of 6 equals infinity end style      

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.

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