Pertanyaan

Nilai maksimum dari adalah ….

Nilai maksimum dari $f open parentheses x close parentheses equals fraction numerator 8 cos invisible function application x plus 6 sin invisible function application x plus 2 over denominator 4 cos invisible function application x plus 3 sin invisible function application x plus 7 end fraction$ adalah ….

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Habis dalam

Klaim

N. Ayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Padang

Jawaban terverifikasi

Pembahasan

Agar diperoleh nilai maksimum f(x) maka tentukan f'(x) terlebih dahulu. Dengan Agar f(x) maksimum maka f'(x) sehingga Maka dan sehingga

$f open parentheses x close parentheses equals fraction numerator u left parenthesis x right parenthesis over denominator v open parentheses x close parentheses end fraction space m a k a space u open parentheses x close parentheses equals 8 cos invisible function application x plus 6 sin invisible function application x plus 2 space d a n space v open parentheses x close parentheses equals 4 cos invisible function application x plus 3 sin invisible function application x plus 7$

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

$f to the power of apostrophe open parentheses x close parentheses equals fraction numerator u to the power of apostrophe open parentheses x close parentheses v open parentheses x close parentheses minus u open parentheses x close parentheses v to the power of apostrophe open parentheses x close parentheses over denominator v squared open parentheses x close parentheses end fraction$

Dengan

$u open parentheses x close parentheses equals 8 cos invisible function application x plus 6 sin invisible function application x plus 2 rightwards arrow u to the power of apostrophe open parentheses x close parentheses equals negative 8 sin invisible function application x plus 6 cos invisible function application x v open parentheses x close parentheses equals 4 cos invisible function application x plus 3 sin invisible function application x plus 7 rightwards arrow v to the power of apostrophe open parentheses x close parentheses equals negative 4 sin invisible function application x plus 3 cos invisible function application x S e h i n g g a f to the power of apostrophe open parentheses x close parentheses equals fraction numerator open parentheses negative 8 sin invisible function application x plus 6 cos invisible function application x close parentheses open parentheses 4 cos invisible function application x plus 3 sin invisible function application x plus 7 close parentheses minus open parentheses 8 cos invisible function application x plus 6 sin invisible function application x plus 2 close parentheses open parentheses negative 4 sin invisible function application x plus 3 cos invisible function application x close parentheses over denominator open parentheses 4 cos invisible function application x plus 3 sin invisible function application x plus 7 close parentheses squared end fraction f apostrophe open parentheses x close parentheses equals fraction numerator negative 48 sin invisible function application x plus 36 cos invisible function application x over denominator open parentheses 4 cos invisible function application x plus 3 sin invisible function application x plus 7 close parentheses squared end fraction$

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

$rightwards double arrow fraction numerator negative 48 sin invisible function application x plus 36 cos invisible function application x over denominator open parentheses 4 cos invisible function application x plus 3 sin invisible function application x plus 7 close parentheses squared end fraction equals 0 rightwards double arrow negative 48 sin invisible function application x plus 36 cos invisible function application x equals 0 rightwards double arrow negative 48 sin invisible function application x equals negative 36 cos invisible function application x rightwards double arrow tan invisible function application x equals 3 over 4$

Maka dan sehingga

$rightwards double arrow fraction numerator 8 open parentheses 4 over 5 close parentheses plus 6 open parentheses 3 over 5 close parentheses plus 2 over denominator 4 open parentheses 4 over 5 close parentheses plus 3 open parentheses 3 over 5 close parentheses plus 7 end fraction rightwards double arrow fraction numerator 32 over 5 plus 18 over 5 plus 2 over denominator 16 over 5 plus 9 over 5 plus 7 end fraction rightwards double arrow 12 over 12 equals 1$

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