12. Jika $f(p)=p+\frac{1}{x}$ dan $g(p)=p-\frac{1}{p}$, maka $g(f(p))$ adalah …
A. $p^2 - \frac{1}{p^2}$
B. $\frac{P^2 +1}{p} - \frac{p}{p^2 +1}$
C. $\frac{P^2 -1}{p} + \frac{p}{p^2 -1}$
D. $2x$
STIS 2007/14
Jawaban: B
$g(f(p))=g(p+\frac{1}{p})$
$ g(p+\frac{1}{p})=g(\frac{p^2+1}{x})$
$ g(\frac{p^2+1}{x})=\frac{p^2+1}{p}-\frac{1}{\frac{p^2+1}{p}}$
$ g(\frac{p^2+1}{x})=\ frac{p^2+1}{p}-\frac{p}{p^2+1}$
PEMBAHASAN SOAL FUNGSI KOMPOSISI DAN INVERS TAHUN 2007 NO 12 STIS
Jawaban: B
$g(f(p))=g(p+\frac{1}{p})$
$ g(p+\frac{1}{p})=g(\frac{p^2+1}{x})$
$ g(\frac{p^2+1}{x})=\frac{p^2+1}{p}-\frac{1}{\frac{p^2+1}{p}}$
$ g(\frac{p^2+1}{x})=\ frac{p^2+1}{p}-\frac{p}{p^2+1}$
PEMBAHASAN SOAL FUNGSI KOMPOSISI DAN INVERS TAHUN 2007 NO 12 STIS
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