Contoh Soal 12 Turunan Fungsi Trigonometri (Bagian 3)

$\begin{array}{ll}\\ 56.&\textrm{Diketahui fungsi}\: \: f(x)=\displaystyle \frac{1}{2}\sin 2x\: \: \textrm{dengan}\\ &0^{\circ}<x<360^{\circ} \: .\: \textrm{Kurva akan cekung}\\ &\textrm{ke atas pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0^{\circ}<x<90^{\circ}\\ \textrm{b}.&0^{\circ}<x<90^{\circ}\: \: \textrm{atau}\: \: 180^{\circ}<x<270^{\circ}\\ \textrm{c}.&45^{\circ}<x<225^{\circ}\\ \color{red}\textrm{d}.&90^{\circ}<x<180^{\circ}\: \: \textrm{atau}\: \: 270^{\circ}<x<360^{\circ}\\ \textrm{e}.&180^{\circ}<x<225^{\circ}\: \: \textrm{atau}\: \: 225^{\circ}<x<360^{\circ} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&f(x)=\displaystyle \frac{1}{2}\sin 2x\\ &f'(x)=\cos 2x\Rightarrow f''(x)=-2\sin 2x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &-2\sin 2x=0\Leftrightarrow \sin 2x=0\\ &\Leftrightarrow \sin 2x=\sin 0^{\circ}\\ &\Leftrightarrow 2x=0^{\circ}+k.360^{\circ}\: \: \textrm{atau}\: \: 2x=180^{\circ}+k.360^{\circ}\\ &\Leftrightarrow x=0^{\circ}+k.180^{\circ}\: \: \textrm{atau}\: \: x=90^{\circ}+k.180^{\circ}\\ &\Leftrightarrow \color{red}x=0^{\circ},\: x=90^{\circ} \: ,\: x=180^{\circ}\: \: \textrm{dan}\: \: x=270^{\circ}\\ &\qquad \color{red}\textrm{serta}\: \: x=360^{\circ}\\ &\bullet \color{red}\textrm{Selang}\: \: 0^{\circ}<x<90^{\circ},\: \: \color{black}\textrm{misal}\: \: x=45^{\circ}\\ &\Rightarrow \Rightarrow f''=-2\sin 2\left ( 45^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: 90^{\circ}<x<180^{\circ},\: \: \color{black}\textrm{misal}\: \: x=135^{\circ}\\ &\Rightarrow \Rightarrow f''=-2\sin 2\left ( 135^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas}\\ &\bullet \color{red}\textrm{Selang}\: \: 180^{\circ}<x<270^{\circ},\: \: \color{black}\textrm{misal}\: \: x=225^{\circ}\\ &\Rightarrow \Rightarrow f''=-2\sin 2\left ( 225^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: 270^{\circ}<x<360^{\circ},\: \: \color{black}\textrm{misal}\: \: x=315^{\circ}\\ &\Rightarrow \Rightarrow f''=-2\sin 2\left ( 315^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 57.&\textrm{Diketahui fungsi}\: \: f(x)=\cos ^{2}x-\sin ^{2}x\: \: \textrm{dengan}\\ &0<x<2\pi \: .\: \textrm{Kurva akan cekung ke bawah}\\ &\textrm{pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0<x<\displaystyle \frac{\pi }{2}\\ \textrm{b}.&\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4}\\ \color{red}\textrm{c}.&\displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{7\pi }{4}<x<2\pi \\ \textrm{d}.&\displaystyle \frac{7\pi }{4}<x<2\pi \\ \textrm{e}.&\displaystyle \frac{5\pi }{4}<x<2\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&f(x)=\color{red}\cos ^{2}x-\sin ^{2}x\color{blue}=\cos 2x\\ &f'(x)=-2\sin 2x\Rightarrow f''(x)=-4\cos 2x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &-4\cos 2x=0\Leftrightarrow \cos 2x=0\\ &\Leftrightarrow \cos 2x=\cos \displaystyle \frac{\pi }{2}\\ &\Leftrightarrow 2x=\pm \displaystyle \frac{\pi }{2}+k.2\pi \: \: \Leftrightarrow \: \: x=\displaystyle \frac{\pi }{4}+k.\pi \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{4},\: x=\frac{3\pi }{4} \: ,\: x=\frac{5\pi }{4}\: \: \color{black}\textrm{dan}\: \: \color{red}x=\frac{7\pi }{4}\\ &\qquad \color{red}\textrm{Ingat bahwa domain}\: \: 0<x<2\pi \: \: \textrm{saja}\\ &\bullet \color{red}\textrm{Selang}\: \: 0<x<\displaystyle \frac{\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=30^{\circ}=\frac{\pi }{6}\\ &\Rightarrow f''(30^{\circ})=-4\cos 2\left ( 30^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{\pi }{4}<x<\displaystyle \frac{3\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=120^{\circ}=\displaystyle \frac{2\pi }{3}\\ &\Rightarrow f''(120^{\circ})=-4\cos 2\left ( 90^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=210^{\circ}=\frac{7\pi }{6}\\ &\Rightarrow f''(210^{\circ})=-4\cos 2\left ( 210^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=300^{\circ}=\displaystyle \frac{5\pi }{3}\\ &\Rightarrow f''(300^{\circ})=-4\cos 2\left ( 300^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{7\pi }{4}<x<2\pi ,\: \: \color{black}\textrm{misal}\: \: x=330^{\circ}=\frac{11\pi }{6}\\ &\Rightarrow f''=-4\cos 2\left ( 330^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 58.&\textrm{Diketahui fungsi}\: \: f(x)=\sin ^{2}x\: \: \textrm{dengan}\\ &0<x<2\pi .\: \textrm{Kurva fungsi tersebut akan}\\ &\textrm{cekung ke bawah pada interval}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\: \: \textrm{atau}\: \: \frac{5\pi }{4}<x<\frac{7\pi }{4}\\ \textrm{b}.&\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\: \: \textrm{atau}\: \: \frac{7\pi }{4}<x<2\pi\\ \textrm{c}.&\displaystyle 0<x<\frac{\pi }{2}\: \: \textrm{atau}\: \: \frac{3\pi }{4}<x<\frac{5\pi }{4}\\ \textrm{d}.&\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\\ \textrm{e}.&\displaystyle 0<x<\frac{\pi }{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{black}\begin{aligned}&f(x)=\color{red}\sin ^{2}x\\ &f'(x)=2\sin x\cos x=\sin 2x\\ & f''(x)=2\cos 2x\\ &\color{purple}\textrm{Syarat belok}\: \: f''(x)=0\\ &2\cos 2x=0\Leftrightarrow \cos 2x=0\\ &\Leftrightarrow \cos 2x=\cos \displaystyle \frac{\pi }{2}\\ &\Leftrightarrow 2x=\pm \displaystyle \frac{\pi }{2}+k.2\pi \: \: \Leftrightarrow \: \: x=\displaystyle \frac{\pi }{4}+k.\pi \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{4},\: x=\frac{3\pi }{4} \: ,\: x=\frac{5\pi }{4}\: \: \color{black}\textrm{dan}\: \: \color{red}x=\frac{7\pi }{4}\\ &\qquad \color{red}\textrm{Ingat bahwa domain}\: \: 0<x<2\pi \: \: \textrm{saja}\\ &\bullet \color{red}\textrm{Selang}\: \: 0<x<\displaystyle \frac{\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=30^{\circ}=\frac{\pi }{6}\\ &\Rightarrow f''(30^{\circ})=2\cos 2\left ( 30^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{\pi }{4}<x<\displaystyle \frac{3\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=120^{\circ}=\displaystyle \frac{2\pi }{3}\\ &\Rightarrow f''(120^{\circ})=2\cos 2\left ( 90^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=210^{\circ}=\frac{7\pi }{6}\\ &\Rightarrow f''(210^{\circ})=2\cos 2\left ( 210^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=300^{\circ}=\displaystyle \frac{5\pi }{3}\\ &\Rightarrow f''(300^{\circ})=2\cos 2\left ( 300^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{7\pi }{4}<x<2\pi ,\: \: \color{black}\textrm{misal}\: \: x=330^{\circ}=\frac{11\pi }{6}\\ &\Rightarrow f''=2\cos 2\left ( 330^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 59.&\textrm{Diketahui fungsi}\: \: f(x)=2\sin x-2\cos x\: \: \textrm{dengan}\\ &0<x<2\pi \: .\: \textrm{Kurva akan cekung ke atas}\\ &\textrm{pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0<x<\displaystyle \frac{3\pi }{4}\\ \textrm{b}.&\displaystyle \frac{\pi }{4}<x <\displaystyle \frac{5\pi }{4}\\ \textrm{c}.& \displaystyle \frac{3\pi }{4}<x<2\pi \\ \textrm{d}.&0<x<\frac{\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4}\\ \color{red}\textrm{e}.&0<x<\frac{\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{5\pi }{4}<x<2\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&f(x)=\color{red}2\sin x-2\cos x\color{blue}\\ &f'(x)=2\cos x+2\sin x\\ & f''(x)=-2\sin x+2\cos x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &-2\sin x+2\cos x=0\Leftrightarrow \sin x=\cos x\\ &\Leftrightarrow \tan x=1\\ &\Leftrightarrow x=\displaystyle \frac{\pi }{4}+k.\pi \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{4},\: x=\frac{5\pi }{4}\\ &\qquad \color{red}\textrm{Ingat bahwa domain}\: \: 0<x<2\pi \: \: \textrm{saja}\\ &\color{black}\textrm{Sebagai gambaran saja}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{\pi }{4}<x<\displaystyle \frac{3\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=90^{\circ}=\displaystyle \frac{\pi }{2}\\ &\Rightarrow f''(90^{\circ})=-2\sin 90^{\circ}+2\cos 90^{\circ}=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 60.&\textrm{Diketahui fungsi}\: \: f(x)=\sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )\\ &\textrm{dengan}\: \: 0<x<2\pi .\: \textrm{Kurva fungsi tersebut}\\ &\textrm{akan cekung ke atas pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 0<x<\frac{\pi }{6}\: \: \textrm{atau}\: \: \frac{\pi }{2}<x<\frac{5\pi }{6}\\ \color{red}\textrm{b}.&\color{magenta}\displaystyle \frac{\pi }{6}<x<\frac{\pi }{2}\: \: \textrm{atau}\: \: \frac{5\pi }{6}<x<\pi\\ \textrm{c}.&\displaystyle \frac{\pi }{6}<x<\frac{\pi }{2}\: \: \textrm{atau}\: \: \frac{3\pi }{4}<x<\frac{5\pi }{6}\\ \textrm{d}.&\displaystyle \frac{\pi }{6}<x<\frac{\pi }{4}\: \: \textrm{atau}\: \: \frac{3\pi }{4}<x<\frac{5\pi }{6}\\ \textrm{e}.&\displaystyle \frac{\pi }{6}<x<\frac{\pi }{4}\: \: \textrm{atau}\: \: \frac{5\pi }{6}<x<\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&f(x)=\sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )\\ &f'(x)=3\cos \left ( 3x+\displaystyle \frac{\pi }{2} \right )\\ & f''(x)=-9\sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &-9\sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )=0\Leftrightarrow \sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )=0\\ &\Leftrightarrow \sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )=\sin 0\\ &\Leftrightarrow \left ( 3x+\displaystyle \frac{\pi }{2} \right )=0+k.2\pi \: \: \Leftrightarrow \: \: \left ( 3x+\displaystyle \frac{\pi }{2} \right )=\displaystyle \pi+k.2\pi \\ &\Leftrightarrow 3x=-\displaystyle \frac{\pi }{2}+k.2\pi \: \: \Leftrightarrow \: \: 3x=\displaystyle \frac{\pi }{2}+k.2\pi \\ &\Leftrightarrow x=-\displaystyle \frac{\pi }{6}+k.\frac{2\pi}{3} \: \: \Leftrightarrow \: \: x=\displaystyle \frac{\pi }{6}+k.\displaystyle \frac{2\pi}{3} \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{6},\: \: x=\displaystyle \frac{\pi }{2},\: x=\frac{5\pi }{6} \: ,\: x=\frac{7\pi }{6},\\ & \color{black}\textrm{dan}\: \: \color{red}x=\frac{3\pi }{2},\: \color{black}\textrm{serta}\: \: \color{red}x=\frac{11\pi}{6} \\ &\qquad \color{red}\textrm{Ingat bahwa domain}\: \: 0\leq x\leq 2\pi \: \: \textrm{saja}\\ &\color{black}\textrm{Sebagai GAMBARAN saja, diberikan 2 nilai selang}\\ &\bullet \color{red}\textrm{Selang}\: \: 0<x<\displaystyle \frac{\pi }{6},\: \: \color{black}\textrm{misal}\: \: x=15^{\circ}=\frac{\pi }{12}\\ &\Rightarrow f''(15^{\circ})=-9\sin \left ( 3\left ( \displaystyle \frac{\pi }{12} \right )+\displaystyle \frac{\pi }{2} \right )=-\displaystyle \frac{9}{2}\sqrt{2}<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{\pi }{6}<x<\displaystyle \frac{\pi }{2},\: \: \color{black}\textrm{misal}\: \: x=60^{\circ}=\displaystyle \frac{\pi }{3}\\ &\Rightarrow f''(60^{\circ})=-9\sin \left ( 3\left ( \displaystyle \frac{\pi }{3} \right )+\displaystyle \frac{\pi }{2} \right )=9>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas} \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan MAtematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
2. Tasari, Aksin, N., Miyanto, Muklis. 2016. Matematika untuk SMA/MA Kelas XII Peminatan Matematika dan Ilmu-Ilmu Alam. Klaten. PT. INTAN PARIWARA.