Pertanyaan

Nilai maksimum dari f ( x ) = 4 cos x + 3 sin x + 7 8 cos x + 6 sin x + 2 adalah...

Nilai maksimum dari $f (x) = \frac{8 cos x + 6 sin x + 2}{4 cos x + 3 sin x + 7}$ adalah...

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Habis dalam

Klaim

N. Mustikowati

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Pembahasan

maka u(x) = 8 cos x + 6 sin x + 2 dan v(x) = 4 cos x + 3 sin x + 7. Agar diperoleh nilai maksimum f(x) maka tentukan f'(x) terlebih dahulu. Dengan Sehingga Agar f(x) maksimum maka f'(x) = 0 sehingga Maka dan sehingga

$begin mathsize 12px style f open parentheses x close parentheses equals fraction numerator u open parentheses x close parentheses over denominator v open parentheses x close parentheses end fraction end style$ maka u(x) = 8 cos x + 6 sin x + 2 dan v(x) = 4 cos x + 3 sin x + 7.

Agar diperoleh nilai maksimum f(x) maka tentukan f'(x) terlebih dahulu.

$begin mathsize 12px style f apostrophe open parentheses x close parentheses equals fraction numerator u apostrophe open parentheses x close parentheses v open parentheses x close parentheses minus u open parentheses x close parentheses v apostrophe open parentheses x close parentheses over denominator v squared open parentheses x close parentheses end fraction end style$

Dengan

Sehingga

$begin mathsize 12px style f apostrophe open parentheses x close parentheses equals fraction numerator open parentheses negative 8 space sin space x plus 6 space cos space x close parentheses open parentheses 4 space cos space x plus 3 space sin space x plus 7 close parentheses minus open parentheses 8 space cos space x plus 6 space sin space x plus 2 close parentheses open parentheses negative 4 space sin space x plus 3 space cos space x close parentheses over denominator open parentheses 4 space cos space x plus 3 space sin space x plus 7 close parentheses squared end fraction f apostrophe open parentheses x close parentheses equals fraction numerator negative 48 space sin space x plus 36 space cos space x over denominator open parentheses 4 space cos space x plus 3 space sin space x plus 7 close parentheses squared end fraction end style$

Agar f(x) maksimum maka f'(x) = 0 sehingga

$begin mathsize 12px style rightwards double arrow fraction numerator negative 48 space sin space x plus 36 space cos space x over denominator open parentheses 4 space cos space x plus 3 space sin space x plus 7 close parentheses squared end fraction equals 0 rightwards double arrow negative 48 space sin space x plus 36 space cos space x equals 0 rightwards double arrow negative 48 sin space x equals negative 36 space cos space x rightwards double arrow tan space x equals 3 over 4 end style$

Maka dan sehingga

$begin mathsize 12px style f open parentheses x close parentheses equals fraction numerator 8 space cos space x plus 6 space sin space x plus 2 over denominator 4 space cos space x plus 3 space sin space x plus 7 end fraction f open parentheses x close parentheses equals fraction numerator 8 open parentheses begin display style 4 over 5 end style close parentheses plus 6 open parentheses begin display style 3 over 5 end style close parentheses plus 2 over denominator 4 open parentheses 4 over 5 close parentheses plus 3 open parentheses 3 over 5 close parentheses plus 7 end fraction f open parentheses x close parentheses equals fraction numerator begin display style 32 over 5 end style plus begin display style 18 over 5 end style plus 2 over denominator begin display style 16 over 5 end style plus begin display style 9 over 5 end style plus 7 end fraction f open parentheses x close parentheses equals 12 over 12 equals 1 end style$