Diketahui fungsi f(x) = 2 cos (2x − (π/3)) dalam selang [0,2π]. Tentukan selang fungsi naik dan fungsi turunnya!

<p>Jawaban yang benar adalah fungsi naik pada interval x &lt; π/6 atau 2π/3 &lt; x &lt; 7π/6 atau x &gt; 5π/3 dan turun pada interval π/6 &lt; x &lt; 2π/3 atau 7π/6 &lt; x &lt; 5π/3</p><p>&nbsp;</p><p>Ingat kembali:&nbsp;</p><p>1. Fungsi f(x) naik dalam suatu interval jika f'(x) &gt; 0 untuk setiap x anggota interval tersebut.&nbsp;</p><p>2. Fungsi f(x) turun dalam suatu interval jika f'(x) &lt; 0 untuk setiap x anggota interval tersebut.&nbsp;</p><p>3. Rumus turunan fungsi trigonometri: f(x) = cos (ax - b) maka f'(x) = -a sin (ax - b)&nbsp;</p><p>4. Persamaan trigonometri untuk sinus:&nbsp;</p><p>sin x = sin a&nbsp;</p><p>x₁ = a + k·2π</p><p>x₂ = (π - a) + k·2π</p><p>&nbsp;</p><p>Pembahasan:&nbsp;</p><p>f(x) = 2 cos {2x - (π/3)}&nbsp;</p><p>f'(x) = 2·[- 2 sin {2x - (π/3)}]&nbsp;</p><p>f'(x) = -4 sin {2x - (π/3)} dalam selang [0, 2π]&nbsp;</p><p>&nbsp;</p><p>Pembuat nol:&nbsp;</p><p>-4 sin {2x - (π/3)} = 0&nbsp;</p><p>sin {2x - (π/3)} = 0/(-4)&nbsp;</p><p>sin {2x - (π/3)} = 0&nbsp;</p><p>sin {2x - (π/3)} = sin 0&nbsp;</p><p>2x₁ - (π/3) = 0 + k·2π</p><p>2x₁ = {0 + (π/3)} + k·2π</p><p>2x₁ = (π/3) + k·2π</p><p>x₁ = {(π/3)/2} + (k·2π)/2</p><p>x₁ = (π/6) + k·π</p><p>k = 0</p><p>x₁ = (π/6) + 0·π</p><p>x₁ = (π/6) + 0</p><p>x₁ = π/6 (memenuhi interval)</p><p>k = 1</p><p>x₁ = (π/6) + 1·π</p><p>x₁ = (π/6) + π</p><p>x₁ = 7π/6 (memenuhi interval)</p><p>k = 2</p><p>x₁ = (π/6) + 2·π</p><p>x₁ = (π/6) + 2π</p><p>x₁ = 13π/6 (tidak memenuhi interval)</p><p>&nbsp;</p><p>2x₂ - (π/3) = (π - 0) + k·2π</p><p>2x₂ - (π/3) = π + k·2π</p><p>2x₂= {π + (π/3)} + k·2π</p><p>2x₂ = (4π/3) + k·2π</p><p>x₂ = {(4π/3)/2} + (k·2π)/2</p><p>x₂ = (2π/3) + k·π</p><p>k = 0</p><p>x₂ = (2π/3) + 0·π</p><p>x₂ = (2π/3) + 0</p><p>x₂ = 2π/3 (memenuhi interval)</p><p>k = 1</p><p>x₂ = (2π/3) + 1·π</p><p>x₂ = (2π/3) + π</p><p>x₂ = 5π/3 (memenuhi interval)</p><p>k = 2</p><p>x₂ = (2π/3) + 2·π</p><p>x₂ = (2π/3) + 2π</p><p>x₂ = 8π/3 (tidak memenuhi interval)</p><p>&nbsp;</p><p>Uji titik 0 untuk interval x &lt; π/6</p><p>-4 sin {2x - (π/3)}&nbsp;</p><p>= -4 sin {2·(0) - (π/3)}&nbsp;</p><p>= -4 sin {0 - (π/3)}</p><p>= -4 sin (-π/3)</p><p>= -4·(-½√3)</p><p>= 2√3 (positif)</p><p>&nbsp;</p><p>Uji titik π/2 untuk interval π/6 &lt; x &lt; 2π/3</p><p>-4 sin {2x - (π/3)}&nbsp;</p><p>= -4 sin {2·(π/2) - (π/3)}</p><p>= -4 sin {π - (π/3)}</p><p>= -4 sin (2π/3)</p><p>= -4·(½√3)</p><p>= -2√3 (negatif)</p><p>&nbsp;</p><p>Uji titik π untuk interval 2π/3 &lt; x &lt; 7π/6</p><p>-4 sin {2x - (π/3)}&nbsp;</p><p>= -4 sin {2·π - (π/3)}</p><p>= -4 sin (2π - (π/3)</p><p>= -4 sin (5π/3)</p><p>= -4·(-½√3)</p><p>= 2√3 (positif)</p><p>&nbsp;</p><p>Uji titik 3π/2 untuk interval 7π/6 &lt; x &lt; 5π/3</p><p>-4 sin {2x - (π/3)}&nbsp;</p><p>= -4 sin {2·(3π/2) - (π/3)}</p><p>= -4 sin {3π - (π/3)}</p><p>= -4 sin (8π/3)</p><p>= -4·(½√3)</p><p>= -2√3 (negatif)</p><p>&nbsp;</p><p>Uji titik 2π untuk interval x &gt; 5π/3</p><p>-4 sin {2x - (π/3)}&nbsp;</p><p>= -4 sin {2·2π - (π/3)}</p><p>= -4 sin {4π - (π/3)}</p><p>= -4 sin (11π/3)</p><p>= -4·(-½√3)</p><p>= 2√3 (positif)</p><p>&nbsp;</p><p>++ (π/6) ---- (2π/3) ++ (7π/6) ---- (5π/3) ++</p><p>––––o–––––––o––––––––o––––––––o––––</p><p>&nbsp;</p><p>Fungsi naik jika f'(x) &gt; 0&nbsp;</p><p>➡️ ambil yang bertanda positif yaitu:</p><p>x &lt; π/6 atau 2π/3 &lt; x &lt; 7π/6 atau x &gt; 5π/3</p><p>&nbsp;</p><p>Fungsi turun jika f'(x) &lt; 0&nbsp;</p><p>➡️ ambil yang bertanda negatif yaitu:</p><p>π/6 &lt; x &lt; 2π/3 atau 7π/6 &lt; x &lt; 5π/3</p><p>&nbsp;</p><p>Jadi, fungsi naik pada interval x &lt; π/6 atau 2π/3 &lt; x &lt; 7π/6 atau x &gt; 5π/3 dan turun pada interval π/6 &lt; x &lt; 2π/3 atau 7π/6 &lt; x &lt; 5π/3</p>