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Fungsi turun pada interval ...

$begin mathsize 14px style 0 less than x less than fraction numerator 5 pi over denominator 12 end fraction end style$
$fraction numerator size 14px 5 size 14px pi over denominator size 14px 12 end fraction size 14px less than size 14px x size 14px less than fraction numerator size 14px 7 size 14px pi over denominator size 14px 12 end fraction$
$size 14px minus fraction numerator size 14px 7 size 14px pi over denominator size 14px 12 end fraction size 14px less than size 14px x size 14px less than size 14px pi over size 14px 12$

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A. Rizky

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Mahasiswa/Alumni Universitas Negeri Malang

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$begin mathsize 14px style text i) Tentukan turunan pertama dari fungsi end text f open parentheses x close parentheses equals negative square root of cos squared x plus x over 2 plus pi end root f open parentheses x close parentheses equals negative open parentheses cos squared x plus x over 2 plus pi close parentheses to the power of 1 half end exponent f to the power of apostrophe open parentheses x close parentheses equals negative 1 half open parentheses cos squared x plus x over 2 plus pi close parentheses to the power of negative 1 half end exponent bullet open parentheses 2 cos space x open parentheses negative sin space x close parentheses plus 1 half close parentheses f to the power of apostrophe open parentheses x close parentheses equals negative fraction numerator negative 2 sin space x space cos space x plus 1 half over denominator 2 square root of cos squared x plus x over 2 plus pi end root end fraction text ii) Fungsi turun jika turunan pertamanya negatif end text f to the power of apostrophe open parentheses x close parentheses less than 0 minus fraction numerator negative 2 sin space x space cos space x plus 1 half over denominator 2 square root of cos squared x plus x over 2 plus pi end root end fraction less than 0 fraction numerator negative 2 sin space x space cos space x plus 1 half over denominator 2 square root of cos squared space x plus x over 2 plus pi end root end fraction greater than 0 text Karena end text x greater than negative pi comma text maka end text x over 2 plus pi greater than 0 text Dan karena end text cos squared x greater or equal than 0 text , maka end text cos squared x plus x over 2 plus pi greater than 0 text Sehingga end text 2 square root of cos squared x plus x over 2 plus pi end root greater than 0 text , maka: end text minus 2 sin space x space cos space x plus 1 half greater than 0 minus 2 sin space x space cos space x greater than negative 1 half 2 sin space x space cos space x less than 1 half sin space 2 x less than 1 half sin space 2 x minus 1 half less than 0 text iii) Tentukan pembuat nol pada bentuk dari ruas kiri dan tempatkan pada garis bilangan end text sin space 2 x minus 1 half less than 0 text Maka end text sin space 2 x minus 1 half equals 0 sin space 2 x equals 1 half sin space 2 x equals sin pi over 6 text Ingat! Jika end text space sin space x equals sin space alpha comma space text maka end text x equals alpha plus-or-minus 2 pi k space text atau end text x equals open parentheses pi minus alpha close parentheses plus-or-minus 2 pi k text Maka untuk end text sin text end text 2 x equals sin pi over 6 text didapat penyelesaian end text 2 x equals pi over 6 plus-or-minus 2 pi k rightwards arrow x equals pi over 12 plus-or-minus pi k text atau end text 2 x equals open parentheses pi minus pi over 6 close parentheses plus-or-minus 2 pi k rightwards arrow 2 x equals fraction numerator 5 pi over denominator 6 end fraction plus-or-minus 2 pi k rightwards arrow x equals fraction numerator 5 pi over denominator 12 end fraction plus pi k text Untuk selang end text minus pi less than x less than 2 pi space text dipenuhi untuk end text x equals negative 11 over 12 pi comma negative 7 over 12 pi comma 1 over 12 pi comma 5 over 12 pi comma 13 over 12 pi comma 17 over 12 pi. text Pada garis bilangan end text end style$

begin mathsize 14px style text v) Tetapkan daerah penyelesaian end text text Karena pertidaksamaan akhir berbentuk end text sin space 2 x minus 1 half less than 0 text Maka dipilih daerah-daerah yang bernilai (-), yaitu end text end style

begin mathsize 14px style text Sehingga didapat daerah penyelesaian end text minus pi less than x less than negative 11 over 12 pi text atau end text minus 7 over 12 pi less than x less than 1 over 12 pi text atau end text 5 over 12 pi less than x less than 13 over 12 pi text atau end text 17 over 12 pi less than x less than 2 pi end style