Contoh Soal 12 Turunan Fungsi Trigonometri (Bagian 3)

$\begin{array}{l}\\ 56.& \text{Diketahui fungsi}f(x)={\displaystyle \frac{1}{2}\sin 2x\text{dengan}}\\ & 0^{\circ }<x<{360}^{\circ }.\text{Kurva akan cekung}\\ & \text{ke atas pada interval}....\\ & \begin{array}{l}\\ \text{a}.& 0^{\circ }<x<{90}^{\circ }\\ \text{b}.& 0^{\circ }<x<{90}^{\circ }\text{atau}{180}^{\circ }<x<{270}^{\circ }\\ \text{c}.& {45}^{\circ }<x<{225}^{\circ }\\ \text{d}.& {90}^{\circ }<x<{180}^{\circ }\text{atau}{270}^{\circ }<x<{360}^{\circ }\\ \text{e}.& {180}^{\circ }<x<{225}^{\circ }\text{atau}{225}^{\circ }<x<{360}^{\circ }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& f(x)={\displaystyle \frac{1}{2}\sin 2x}\\ & f^{\prime }(x)=\cos 2x\Rightarrow f^{″}(x)=-2\sin 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & -2\sin 2x=0\iff \sin 2x=0\\ & \iff \sin 2x=\sin 0^{\circ }\\ & \iff 2x=0^{\circ }+k{.360}^{\circ }\text{atau}2x={180}^{\circ }+k{.360}^{\circ }\\ & \iff x=0^{\circ }+k{.180}^{\circ }\text{atau}x={90}^{\circ }+k{.180}^{\circ }\\ & \iff x=0^{\circ },x={90}^{\circ },x={180}^{\circ }\text{dan}x={270}^{\circ }\\ & \text{serta}x={360}^{\circ }\\ & \bullet \text{Selang}{0}^{\circ }<x<{90}^{\circ },\text{misal}x={45}^{\circ }\\ & \Rightarrow \Rightarrow f^{″}=-2\sin 2\left({45}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{90}^{\circ }<x<{180}^{\circ },\text{misal}x={135}^{\circ }\\ & \Rightarrow \Rightarrow f^{″}=-2\sin 2\left({135}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\\ & \bullet \text{Selang}{180}^{\circ }<x<{270}^{\circ },\text{misal}x={225}^{\circ }\\ & \Rightarrow \Rightarrow f^{″}=-2\sin 2\left({225}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{270}^{\circ }<x<{360}^{\circ },\text{misal}x={315}^{\circ }\\ & \Rightarrow \Rightarrow f^{″}=-2\sin 2\left({315}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\end{array}\end{array}$

$\begin{array}{l}\\ 57.& \text{Diketahui fungsi}f(x)={\cos }^{2}x-{\sin }^{2}x\text{dengan}\\ & 0<x<2\pi .\text{Kurva akan cekung ke bawah}\\ & \text{pada interval}....\\ & \begin{array}{l}\\ \text{a}.& 0<x<{\displaystyle \frac{\pi }{2}}\\ \text{b}.& {\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\text{atau}{\displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4}}}\\ \text{c}.& {\displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4}\text{atau}{\displaystyle \frac{7\pi }{4}<x<2\pi }}\\ \text{d}.& {\displaystyle \frac{7\pi }{4}<x<2\pi }\\ \text{e}.& {\displaystyle \frac{5\pi }{4}<x<2\pi }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& f(x)={\cos }^{2}x-{\sin }^{2}x=\cos 2x\\ & f^{\prime }(x)=-2\sin 2x\Rightarrow f^{″}(x)=-4\cos 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & -4\cos 2x=0\iff \cos 2x=0\\ & \iff \cos 2x=\cos {\displaystyle \frac{\pi }{2}}\\ & \iff 2x=\pm {\displaystyle \frac{\pi }{2}+k.2\pi \iff x={\displaystyle \frac{\pi }{4}+k.\pi }}\\ & \iff x={\displaystyle \frac{\pi }{4},x=\frac{3\pi }{4},x=\frac{5\pi }{4}\text{dan}x=\frac{7\pi }{4}}\\ & \text{Ingat bahwa domain}0<x<2\pi \text{saja}\\ & \bullet \text{Selang}0<x<{\displaystyle \frac{\pi }{4},\text{misal}x={30}^{\circ }=\frac{\pi }{6}}\\ & \Rightarrow f^{″}({30}^{\circ })=-4\cos 2\left({30}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{\displaystyle \frac{\pi }{4}<x<{\displaystyle \frac{3\pi }{4},\text{misal}x={120}^{\circ }={\displaystyle \frac{2\pi }{3}}}}\\ & \Rightarrow f^{″}({120}^{\circ })=-4\cos 2\left({90}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\\ & \bullet \text{Selang}{\displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4},\text{misal}x={210}^{\circ }=\frac{7\pi }{6}}\\ & \Rightarrow f^{″}({210}^{\circ })=-4\cos 2\left({210}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{\displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4},\text{misal}x={300}^{\circ }={\displaystyle \frac{5\pi }{3}}}\\ & \Rightarrow f^{″}({300}^{\circ })=-4\cos 2\left({300}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\\ & \bullet \text{Selang}{\displaystyle \frac{7\pi }{4}<x<2\pi ,\text{misal}x={330}^{\circ }=\frac{11\pi }{6}}\\ & \Rightarrow f^{″}=-4\cos 2\left({330}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\end{array}\end{array}$

$\begin{array}{l}\\ 58.& \text{Diketahui fungsi}f(x)={\sin }^{2}x\text{dengan}\\ & 0<x<2\pi .\text{Kurva fungsi tersebut akan}\\ & \text{cekung ke bawah pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\text{atau}\frac{5\pi }{4}<x<\frac{7\pi }{4}}\\ \text{b}.& {\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\text{atau}\frac{7\pi }{4}<x<2\pi }\\ \text{c}.& {\displaystyle 0<x<\frac{\pi }{2}\text{atau}\frac{3\pi }{4}<x<\frac{5\pi }{4}}\\ \text{d}.& {\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}}\\ \text{e}.& {\displaystyle 0<x<\frac{\pi }{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& f(x)={\sin }^{2}x\\ & f^{\prime }(x)=2\sin x\cos x=\sin 2x\\ & f^{″}(x)=2\cos 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & 2\cos 2x=0\iff \cos 2x=0\\ & \iff \cos 2x=\cos {\displaystyle \frac{\pi }{2}}\\ & \iff 2x=\pm {\displaystyle \frac{\pi }{2}+k.2\pi \iff x={\displaystyle \frac{\pi }{4}+k.\pi }}\\ & \iff x={\displaystyle \frac{\pi }{4},x=\frac{3\pi }{4},x=\frac{5\pi }{4}\text{dan}x=\frac{7\pi }{4}}\\ & \text{Ingat bahwa domain}0<x<2\pi \text{saja}\\ & \bullet \text{Selang}0<x<{\displaystyle \frac{\pi }{4},\text{misal}x={30}^{\circ }=\frac{\pi }{6}}\\ & \Rightarrow f^{″}({30}^{\circ })=2\cos 2\left({30}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\\ & \bullet \text{Selang}{\displaystyle \frac{\pi }{4}<x<{\displaystyle \frac{3\pi }{4},\text{misal}x={120}^{\circ }={\displaystyle \frac{2\pi }{3}}}}\\ & \Rightarrow f^{″}({120}^{\circ })=2\cos 2\left({90}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{\displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4},\text{misal}x={210}^{\circ }=\frac{7\pi }{6}}\\ & \Rightarrow f^{″}({210}^{\circ })=2\cos 2\left({210}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\\ & \bullet \text{Selang}{\displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4},\text{misal}x={300}^{\circ }={\displaystyle \frac{5\pi }{3}}}\\ & \Rightarrow f^{″}({300}^{\circ })=2\cos 2\left({300}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{\displaystyle \frac{7\pi }{4}<x<2\pi ,\text{misal}x={330}^{\circ }=\frac{11\pi }{6}}\\ & \Rightarrow f^{″}=2\cos 2\left({330}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\end{array}\end{array}$

$\begin{array}{l}\\ 59.& \text{Diketahui fungsi}f(x)=2\sin x-2\cos x\text{dengan}\\ & 0<x<2\pi .\text{Kurva akan cekung ke atas}\\ & \text{pada interval}....\\ & \begin{array}{l}\\ \text{a}.& 0<x<{\displaystyle \frac{3\pi }{4}}\\ \text{b}.& {\displaystyle \frac{\pi }{4}<x<{\displaystyle \frac{5\pi }{4}}}\\ \text{c}.& {\displaystyle \frac{3\pi }{4}<x<2\pi }\\ \text{d}.& 0<x<\frac{\pi }{4}\text{atau}{\displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4}}\\ \text{e}.& 0<x<\frac{\pi }{4}\text{atau}{\displaystyle \frac{5\pi }{4}<x<2\pi }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& f(x)=2\sin x-2\cos x\\ & f^{\prime }(x)=2\cos x+2\sin x\\ & f^{″}(x)=-2\sin x+2\cos x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & -2\sin x+2\cos x=0\iff \sin x=\cos x\\ & \iff \tan x=1\\ & \iff x={\displaystyle \frac{\pi }{4}+k.\pi }\\ & \iff x={\displaystyle \frac{\pi }{4},x=\frac{5\pi }{4}}\\ & \text{Ingat bahwa domain}0<x<2\pi \text{saja}\\ & \text{Sebagai gambaran saja}\\ & \bullet \text{Selang}{\displaystyle \frac{\pi }{4}<x<{\displaystyle \frac{3\pi }{4},\text{misal}x={90}^{\circ }={\displaystyle \frac{\pi }{2}}}}\\ & \Rightarrow f^{″}({90}^{\circ })=-2\sin {90}^{\circ }+2\cos {90}^{\circ }=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\end{array}\end{array}$

$\begin{array}{l}\\ 60.& \text{Diketahui fungsi}f(x)=\sin (3x+{\displaystyle \frac{\pi }{2}})\\ & \text{dengan}0<x<2\pi .\text{Kurva fungsi tersebut}\\ & \text{akan cekung ke atas pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0<x<\frac{\pi }{6}\text{atau}\frac{\pi }{2}<x<\frac{5\pi }{6}}\\ \text{b}.& {\displaystyle \frac{\pi }{6}<x<\frac{\pi }{2}\text{atau}\frac{5\pi }{6}<x<\pi }\\ \text{c}.& {\displaystyle \frac{\pi }{6}<x<\frac{\pi }{2}\text{atau}\frac{3\pi }{4}<x<\frac{5\pi }{6}}\\ \text{d}.& {\displaystyle \frac{\pi }{6}<x<\frac{\pi }{4}\text{atau}\frac{3\pi }{4}<x<\frac{5\pi }{6}}\\ \text{e}.& {\displaystyle \frac{\pi }{6}<x<\frac{\pi }{4}\text{atau}\frac{5\pi }{6}<x<\pi }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& f(x)=\sin (3x+{\displaystyle \frac{\pi }{2}})\\ & f^{\prime }(x)=3\cos (3x+{\displaystyle \frac{\pi }{2}})\\ & f^{″}(x)=-9\sin (3x+{\displaystyle \frac{\pi }{2}})\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & -9\sin (3x+{\displaystyle \frac{\pi }{2}})=0\iff \sin (3x+{\displaystyle \frac{\pi }{2}})=0\\ & \iff \sin (3x+{\displaystyle \frac{\pi }{2}})=\sin 0\\ & \iff (3x+{\displaystyle \frac{\pi }{2}})=0+k.2\pi \iff (3x+{\displaystyle \frac{\pi }{2}})={\displaystyle \pi +k.2\pi }\\ & \iff 3x=-{\displaystyle \frac{\pi }{2}+k.2\pi \iff 3x={\displaystyle \frac{\pi }{2}+k.2\pi }}\\ & \iff x=-{\displaystyle \frac{\pi }{6}+k.\frac{2\pi }{3}\iff x={\displaystyle \frac{\pi }{6}+k.{\displaystyle \frac{2\pi }{3}}}}\\ & \iff x={\displaystyle \frac{\pi }{6},x={\displaystyle \frac{\pi }{2},x=\frac{5\pi }{6},x=\frac{7\pi }{6},}}\\ & \text{dan}x=\frac{3\pi }{2},\text{serta}x=\frac{11\pi }{6}\\ & \text{Ingat bahwa domain}0\le x\le 2\pi \text{saja}\\ & \text{Sebagai GAMBARAN saja, diberikan 2 nilai selang}\\ & \bullet \text{Selang}0<x<{\displaystyle \frac{\pi }{6},\text{misal}x={15}^{\circ }=\frac{\pi }{12}}\\ & \Rightarrow f^{″}({15}^{\circ })=-9\sin (3\left({\displaystyle \frac{\pi }{12}}\right)+{\displaystyle \frac{\pi }{2}})=-{\displaystyle \frac{9}{2}\sqrt{2}<0}\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{\displaystyle \frac{\pi }{6}<x<{\displaystyle \frac{\pi }{2},\text{misal}x={60}^{\circ }={\displaystyle \frac{\pi }{3}}}}\\ & \Rightarrow f^{″}({60}^{\circ })=-9\sin (3\left({\displaystyle \frac{\pi }{3}}\right)+{\displaystyle \frac{\pi }{2}})=9>0\\ & \text{pada selang ini kurva cekung ke atas}\end{array}\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.