Kuis (medium-susah):
Diketahui fungsi berikut ini:
[tex]\displaystyle\rm f(x) = \frac{x^2}{\sqrt{x}-x^3}\\\\Bentuk~~rasional~~=\frac{x^a+x^b}{1-x^c}\\\\Asimtot~~datarnya~~y=p\\Asimtot~~tegaknya~~x=q[/tex]
Nilai ab + c - pq = .....![]()
Diketahui fungsi berikut ini:
[tex]\displaystyle\rm f(x) = \frac{x^2}{\sqrt{x}-x^3}\\\\Bentuk~~rasional~~=\frac{x^a+x^b}{1-x^c}\\\\Asimtot~~datarnya~~y=p\\Asimtot~~tegaknya~~x=q[/tex]
Nilai ab + c - pq = .....
Jawaban:
11
Penjelasan dengan langkah-langkah:
[tex]f(x) = \frac{x^2}{\sqrt{x}-x^3} \times \frac{\sqrt{x} + x^3}{\sqrt{x} + x^3} \\ = \frac{ {x}^{2} \sqrt{x} + {x}^{5} }{x - {x}^{6} } \\ = \frac{x(x \sqrt{x} + {x}^{4} )}{x(1 - {x}^{5} )} \\ = \frac{x \sqrt{x} + {x}^{4}}{1 - {x}^{5}} \\ = \frac{ {x}^{ \frac{3}{2} } + {x}^{4}}{1 - {x}^{5}} [/tex]
[tex]a = \frac{3}{2} \\ b = 4 \\ c = 5[/tex]
[tex]Asimtot \: datar : y = lim_{x - > \infty }( \frac{{x}^{ \frac{3}{2} } + {x}^{4}}{1 - {x}^{5}} ) = 0 \\ Asimtot \: tegak : 1 - {x}^{5} =0 : \: x = 1[/tex]
p = 0
q = 1
[tex]ab + c - pq \\ = (\frac{3}{2} \times 4 )+ 5 - (0 \times 1) \\ = 6 + 5 - 0 \\ = 11[/tex]
[answer.2.content]
