CONTOH SOAL 12 TURUNAN FUNGSI

 

$\begin{array}{ll}\\ 56.&\textrm{Nilai stasioner fungsi}\\ &\quad\quad\quad f(x)=\displaystyle \frac{\sin x}{2-\cos x}\\ &\textrm{untuk}\: \: 0\leq x\leq 2\pi \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left ( \displaystyle \frac{\pi }{2},\frac{1}{2} \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{2},-\frac{1}{2} \right )\\ \textrm{b}.&\left ( \displaystyle \frac{\pi }{3},\frac{1}{2}\sqrt{3} \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{3},-\frac{1}{2}\sqrt{3} \right )\\ \textrm{c}.&\left ( \displaystyle \frac{\pi }{3},\frac{1}{3}\sqrt{3} \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{2\pi }{3},-\frac{1}{3}\sqrt{3} \right )\\ \textrm{d}.&\left ( \displaystyle \frac{\pi }{3},\frac{1}{3}\sqrt{3} \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{5\pi }{3},-\frac{1}{3}\sqrt{3} \right )\\ \textrm{e}.&\left ( \displaystyle \frac{\pi }{4},\frac{1}{4}\sqrt{3} \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{3\pi }{4},-\frac{1}{4}\sqrt{3} \right ) \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Diketahui}\\ & f(x)=\displaystyle \frac{\sin x}{2-\cos x}\Rightarrow f'(x)=\displaystyle \frac{2\cos x-1}{(2-\cos x)^{2}}\\ &\textrm{Stasioner fungsi}\: \: f\: \: \textrm{saat}\: \: f'(x)=0\: \: \textrm{maka},\\ &\displaystyle \frac{2\cos x-1}{(2-\cos x)^{2}}=0\Leftrightarrow 2\cos x-1=0\\ &\Leftrightarrow \cos x=\displaystyle \frac{1}{2}\Leftrightarrow \cos x=\cos \displaystyle \frac{\pi }{3}\\ &\Leftrightarrow x=\pm \displaystyle \frac{\pi }{3}+k.2\pi \\ &\Leftrightarrow k=0\Rightarrow x=\pm \displaystyle \frac{\pi }{3}\Leftrightarrow x=\begin{cases} \displaystyle \frac{\pi }{3} & \textrm{memenuhi} \\ -\displaystyle \frac{\pi }{3} & \textrm{tidak memenuhi} \end{cases}\\ &\Leftrightarrow k=1\Rightarrow x=\pm \displaystyle \frac{\pi }{3}+2\pi \Leftrightarrow x=\begin{cases} \displaystyle \frac{7\pi }{3} & \textrm{tidak memenuhi} \\ \displaystyle \frac{5\pi }{3} & \textrm{memenuhi} \end{cases}\\ &\textrm{Titiknya adalah}\\ &x=\displaystyle \frac{\pi }{3}\Rightarrow f\left ( \displaystyle \frac{\pi }{3} \right )\\ &\qquad=\displaystyle \frac{\sin \displaystyle \frac{\pi }{3}}{2-\cos \displaystyle \frac{\pi }{3}}=\displaystyle \frac{\displaystyle \frac{1}{2}\sqrt{3}}{2-\displaystyle \frac{1}{2}}=\displaystyle \frac{1}{3}\sqrt{3}\\ &\qquad \left ( \displaystyle \frac{\pi }{3},\displaystyle \frac{1}{3}\sqrt{3} \right )\\ &x=\displaystyle \frac{5\pi }{3}\Rightarrow f\left ( \displaystyle \frac{5\pi }{3} \right )\\ &\qquad=\displaystyle \frac{\sin \displaystyle \frac{5\pi }{3}}{2-\cos \displaystyle \frac{5\pi }{3}}=\displaystyle \frac{-\displaystyle \frac{1}{2}\sqrt{3}}{2-\displaystyle \frac{1}{2}}=-\displaystyle \frac{1}{3}\sqrt{3}\\ &\qquad \left ( \displaystyle \frac{5\pi }{3},-\displaystyle \frac{1}{3}\sqrt{3} \right ) \end{aligned} \end{array}$

CONTOH SOAL 11 TURUNAN FUNGSI

 

$\begin{array}{ll}\\ 51.&\textrm{Fungsi}\: \: f(x)=\sqrt{\sin ^{2}x+\displaystyle \frac{x}{2}}\\ & \textrm{dengan}\: \: x\gt 0 \: \: \textrm{turun pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{5\pi }{12}\lt x\leq \frac{13\pi }{12}\\ \textrm{b}.&\displaystyle \frac{7\pi }{12}\lt x\lt \frac{11\pi }{12}\\ \textrm{c}.&\displaystyle \frac{\pi }{12}\lt x\lt \frac{5\pi }{12}\\ \textrm{d}.&\displaystyle \frac{7\pi }{6}\lt x\leq \displaystyle \frac{13\pi }{6} \\ \textrm{e}.&\displaystyle \frac{7\pi }{6}\lt x\leq \frac{11\pi }{6} \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\textrm{Diketahui}\: \: f(x)=\sqrt{\sin ^{2}x+\displaystyle \frac{x}{2}}\\ &\textrm{Fungsi}\: \: f\: \: \textrm{turun, jika}\: \: f'(x)<0\\ &f'(x)=\displaystyle \frac{\sin 2x+\displaystyle \frac{1}{2}}{2\sqrt{\sin^{2} x+\displaystyle \frac{x}{2}}}=0\\ &\sin 2x+\displaystyle \frac{1}{2}=0\Leftrightarrow \sin 2x=-\displaystyle \frac{1}{2}\\ &\Leftrightarrow \sin 2x=\sin \displaystyle \frac{7\pi }{6}\\ &\Leftrightarrow 2x=\displaystyle \frac{7\pi }{6}+ k.2\pi \: \: \textrm{atau}\: \: 2x=\pi -\frac{7\pi }{6}+k.2\pi \\ &\Leftrightarrow x=\displaystyle \frac{7\pi }{12}+ k.\pi \: \: \textrm{atau}\: \: x= -\frac{\pi }{12}+k.\pi \\ &\Leftrightarrow k=0\Rightarrow x=\displaystyle \frac{7\pi }{12}\: \: \textrm{atau}\: \: x=-\displaystyle \frac{\pi }{12}\: \: (\textrm{tm})\\ &\Leftrightarrow k=1\Rightarrow x=\displaystyle \frac{19\pi }{12}\: \: \textrm{atau}\: \: x=\displaystyle \frac{11\pi }{12}\\ &\textrm{Gunakan titik uji pada}\: \: x=\displaystyle \frac{\pi }{2}=90^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'\left ( \displaystyle \frac{\pi }{2} \right )=\displaystyle \frac{\sin 2\left ( \displaystyle \frac{\pi }{2} \right )+\displaystyle \frac{1}{2}}{\sqrt{\sin^{2} \left ( \displaystyle \frac{\pi }{2} \right )+\displaystyle \frac{\left ( \displaystyle \frac{\pi }{2} \right )}{2}}}=+\\ &\qquad (\textrm{positif})\\ &\textrm{Gunakan titik uji pada}\: \: x=\displaystyle \frac{3\pi }{4}=135^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'\left ( \displaystyle \frac{3\pi }{4} \right )=\displaystyle \frac{\sin 2\left ( \displaystyle \frac{3\pi }{4} \right )+\displaystyle \frac{1}{2}}{\sqrt{\sin^{2} \left ( \displaystyle \frac{3\pi }{4} \right )+\displaystyle \frac{\left ( \displaystyle \frac{3\pi }{4} \right )}{2}}}=-\\ &\qquad (\textrm{negatif})\\ &\begin{array}{cccccc}\\ &&&&\\ &++&&--\\\hline 0&&\displaystyle \displaystyle \frac{7\pi }{12}&&\displaystyle \frac{11\pi }{12} \end{array} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 52.&\textrm{Titik stasioner fungsi}\: \: f(x)=\cos 3x\\ & \textrm{pada}\: \: 0\leq x\leq \pi \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&(0,1),\left ( \displaystyle \frac{\pi }{4},1 \right ),\left ( \displaystyle \frac{\pi }{3},1 \right ),\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{2},-1 \right )\\ \textrm{b}.&(0,1),\left ( \displaystyle \frac{\pi }{3},1 \right ),\left ( \displaystyle \frac{\pi }{2},-1 \right ),\: \: \textrm{dan}\: \: \left ( \displaystyle \pi ,-1 \right )\\ \textrm{c}.&\left ( \displaystyle \frac{\pi }{6},-1 \right ),\left ( \displaystyle \frac{\pi }{3},1 \right ),\left ( \displaystyle \frac{\pi }{2},-1 \right ),\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{2\pi }{3},1 \right )\\ \textrm{d}.&\left ( \displaystyle \frac{\pi }{6},1 \right ),\left ( \displaystyle \frac{\pi }{3},-1 \right ),\left ( \displaystyle \frac{\pi }{2},1 \right ),\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{2\pi }{3},-1 \right ) \\ \textrm{e}.&(0,1),\left ( \displaystyle \frac{\pi }{3},-1 \right ),\left ( \displaystyle \frac{2\pi }{3},1 \right ),\: \: \textrm{dan}\: \: \left ( \displaystyle \pi ,-1 \right ) \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll}&\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Diketahui}\\ & f(x)=\cos 3x\Rightarrow f'(x)=-3\sin 3x\\ &\textrm{Stasioner fungsi}\: \: f\: \: \textrm{saat}\: \: f'(x)=0\: \: \textrm{maka},\\ &-\sin 3x=0\Leftrightarrow \sin 3x=0\Leftrightarrow \sin 3x=\sin 0\\ &\Leftrightarrow 3x=0+k.2\pi \: \: \textrm{atau}\: \: 3x=\pi +k.2\pi \\ &\Leftrightarrow x=k.\displaystyle \frac{2\pi }{3}\: \: \textrm{atau}\: \: x=\displaystyle \frac{\pi }{3} +k.\frac{2\pi}{3} \\ &\Leftrightarrow k=0\Rightarrow x=0\: \: \textrm{atau}\: \: x=\displaystyle \frac{\pi }{3}\\ &\Leftrightarrow k=1\Rightarrow x=\displaystyle \frac{2\pi }{3}\: \: \textrm{atau}\: \: x=\displaystyle \pi \\ &\textrm{Sekarang kita tentukan nilai dan titiknya}\\ &x=0\Rightarrow f(0)=\cos 3(0)=1\rightarrow (0,1)\\ &x=\displaystyle \frac{\pi }{3}\Rightarrow f\left ( \displaystyle \frac{\pi }{3} \right )=\cos 3\left ( \displaystyle \frac{\pi }{3} \right )=\cos \pi \\ &\qquad=-1\rightarrow \left ( \displaystyle \frac{\pi }{3},-1 \right )\\ &\textrm{dan seterusnya} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 53.&\textrm{Titik stasioner fungsi}\: \: f(x)=\sin \left ( 2x-\displaystyle \frac{\pi }{6} \right )\\ &\textrm{pada}\: \: 0\leq x\leq \pi \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&(0,1)\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{6},-1 \right )\\ \textrm{b}.&\left ( \displaystyle \frac{\pi }{6},1 \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{3},-1 \right )\\ \textrm{c}.&\left ( \displaystyle \frac{\pi }{4},-1 \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{2},-1 \right )\\ \textrm{d}.&\left ( \displaystyle \frac{\pi }{3},1 \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{5\pi }{6},-1 \right )\\ \textrm{e}.&\left ( \displaystyle \frac{\pi }{2},-1 \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \pi ,1 \right ) \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Diketahui}\\ & f(x)=\sin \left ( 2x-\displaystyle \frac{\pi }{6} \right )\Rightarrow f'(x)=2\cos \left ( 2x-\displaystyle \frac{\pi }{6} \right )\\ &\textrm{Stasioner fungsi}\: \: f\: \: \textrm{saat}\: \: f'(x)=0\: \: \textrm{maka},\\ &2\cos \left ( 2x-\displaystyle \frac{\pi }{6} \right )=0\Leftrightarrow \cos \left ( 2x-\displaystyle \frac{\pi }{6} \right )=0\\ &\cos \left ( 2x-\displaystyle \frac{\pi }{6} \right )=\cos \displaystyle \frac{\pi }{2}\\ &\Leftrightarrow \left ( 2x-\displaystyle \frac{\pi }{6} \right )=\pm \displaystyle \frac{\pi }{2}+k.2\pi \\ &\Leftrightarrow x=\displaystyle \frac{\pi }{12}\pm \frac{\pi }{4}+k.\pi \begin{cases} x & =\displaystyle \frac{\pi }{3}+k.\pi \\ x & =-\displaystyle \frac{\pi }{6}+k.\pi \end{cases}\\ &\Leftrightarrow k=0\Rightarrow \begin{cases} x & =\displaystyle \frac{\pi }{3} \\ x & =-\displaystyle \frac{\pi }{6}\: \: (\textrm{tm}) \end{cases}\\ &\Leftrightarrow k=1\Rightarrow \begin{cases} x & =\displaystyle \frac{4\pi }{3}\: \: \textrm{tm} \\ x & =\displaystyle \frac{5\pi }{6} \end{cases}\\ &\textrm{Sekarang kita tentukan nilai dan titiknya}\\ &x=\displaystyle \frac{\pi }{3}\Rightarrow f\left ( \displaystyle \frac{\pi }{3} \right )=\sin \left (2.\displaystyle \frac{\pi }{3}- \displaystyle \frac{\pi }{6} \right )=\sin \frac{\pi}{2}=1 \\ &\qquad=1\rightarrow \left ( \displaystyle \frac{\pi }{3},1 \right )\\ &x=\displaystyle \frac{5\pi }{6}\Rightarrow f\left ( \displaystyle \frac{5\pi }{6} \right )=\sin \left (2.\displaystyle \frac{5\pi }{6}- \displaystyle \frac{\pi }{6} \right )\\ &\qquad=\sin \frac{3\pi}{2}=-1\rightarrow \left ( \displaystyle \frac{5\pi }{6},-1 \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 54.&\textrm{Nilai}\: \: x\: \: \textrm{pada titik stasioner}\\ &\textrm{fungsi}\: \: f(x)=x+\sin x\: \: \textrm{untuk}\\ &0^{\circ}\leq x\leq 360^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&90^{\circ}\\ \textrm{b}.&135^{\circ}\\ \textrm{c}.&150^{\circ}\\ \textrm{d}.&180^{\circ}\\ \textrm{e}.&360 \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Diketahui}\\ & f(x)=x+\sin x\Rightarrow f'(x)=1+\cos x\\ &\textrm{Stasioner fungsi}\: \: f\: \: \textrm{saat}\: \: f'(x)=0\: \: \textrm{maka},\\ &1+\cos =0\Leftrightarrow \cos x=-1\\ &\Leftrightarrow \cos x=\cos 180^{\circ}\\ &\Leftrightarrow x=\pm 180^{\circ}+k.360^{\circ}\\ &\Leftrightarrow k=0\Rightarrow x=\begin{cases} 180^{\circ} & \textrm{mungkin} \\ -180^{\circ} & \textrm{tidak mungkin} \end{cases}\\ &\Leftrightarrow k=1\Rightarrow x=\begin{cases} 540^{\circ} & \textrm{tidak mungkin} \\ 180^{\circ} & \textrm{mungkin} \end{cases} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 55.&\textrm{Nilai}\: \: y\: \: \textrm{pada titik stasioner}\\ &\textrm{fungsi}\: \: f(x)=4\cos x+\cos 2x\\ &\textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-5\: \: \textrm{dan}\: \: 3\\ \textrm{b}.&-4\: \: \textrm{dan}\: \: 2\\ \textrm{c}.&-3\: \: \textrm{dan}\: \: 5\\ \textrm{d}.&-2\: \: \textrm{dan}\: \: 4\\ \textrm{e}.&3\: \: \textrm{dan}\: \: 5 \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\textrm{Diketahui}\\ & f(x)=4\cos x+\cos 2x\\ &\Rightarrow f'(x)=-4\sin x-2\sin 2x\\ &\textrm{Stasioner fungsi}\: \: f\: \: \textrm{saat}\: \: f'(x)=0\: \: \textrm{maka},\\ &-4\sin x-2\sin 2x=0\\ &\Leftrightarrow -4\sin x-4\sin x\cos x=0\\ &\Leftrightarrow -4\sin x \left ( 1+\cos x \right )=0\\ &\Leftrightarrow \sin x \left ( 1+\cos x \right )=0\\ &\Leftrightarrow \sin x=0\: \: \textrm{atau}\: \: 1+\cos x=0\\ &\Leftrightarrow \sin x=0\: \: \textrm{atau}\: \: \cos x=-1\\ &\Leftrightarrow \sin x=\sin 0^{\circ}\: \: \textrm{atau}\: \: \cos x=\cos 180^{\circ}\\ &\Leftrightarrow x=\begin{cases} 0^{\circ} +k.360^{\circ} \\ 180^{\circ} +k.360^{\circ} \end{cases}\: \textrm{atau}\: \: x=\begin{cases} 180^{\circ} +k.360^{\circ} \\ -180^{\circ} +k.360^{\circ} \end{cases}\\ &\Leftrightarrow k=0\Rightarrow x=0^{\circ}\: \textrm{atau}\: 180^{\circ}\\ &\textrm{Nilai}\: \: y-\textrm{nya}\\ &x=0^{\circ}\Rightarrow f(0^{\circ})\\ &\qquad=4\cos 0^{\circ}+\cos 2(0^{\circ})=4+1=5\\ &x=180^{\circ}\Rightarrow f(180^{\circ})\\ &\qquad=4\cos 180^{\circ}+\cos 2(180^{\circ})=-4+1=-3 \end{aligned} \end{array}$

CONTOH SOAL 10 TURUNAN FUNGSI

 

$\begin{array}{ll}\\ 46.&\textrm{Persamaan garis singgung pada}\\ &\textrm{kurva}\: \: y=\sin x+\cos x\: \: \textrm{di titik yang}\\ &\textrm{berabsis}\: \: \displaystyle \frac{\pi }{2}\: \: \textrm{akan memotong sumbu}\\ &\textrm{Y dengan ordinatnya berupa}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{\pi }{2}+1\\ \textrm{b}.&\displaystyle \frac{\pi }{2}-1\\ \textrm{c}.&1-\displaystyle \frac{\pi }{2}\\ \textrm{d}.&2+\displaystyle \frac{\pi }{2}\\ \textrm{e}.&2-\displaystyle \frac{\pi }{2} \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll}&\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&y=\sin x+\cos x,\: \: \textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{2}\\ &y_{0}=\sin \left (\displaystyle \frac{\pi }{2} \right )+\cos \left ( \displaystyle \frac{\pi }{2} \right )=1+0=1\\ &\textrm{kita cari gradien}\: \: m\: \: \textrm{saat}\: \: y',\: \: \textrm{yaitu}:\\ &m=\cos x-\sin x\\ &m=\cos \left ( \displaystyle \frac{\pi }{2} \right )-\sin \left ( \displaystyle \frac{\pi }{2} \right )\\ &m=0-1=-1\\ &\textrm{Persamaan garis singgungnya adalah}:\\ &y=m(x-x_{0})+y_{0}\\ &\Leftrightarrow \: y=-1\left ( x-\displaystyle \frac{\pi }{2} \right )+1\\ &\Leftrightarrow \: y=-x+\displaystyle \frac{\pi }{2}+1\\ &\textrm{Ordinat garis singgungnya saat}\\ &\textrm{memotong sumbu-Y adalah}:\: \: x=0,\\ &\textrm{maka}\\ &y=-0+\displaystyle \frac{\pi }{2}+1=\displaystyle \frac{\pi }{2}+1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 47.&\textrm{Fungsi}\: \: f(x)=\sin x-\cos x\: \: \textrm{dengan}\\ &0\lt x\lt 2\pi \: \: \textrm{naik pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0\lt x\lt \displaystyle \frac{\pi }{4}\\ \textrm{b}.&\displaystyle \frac{\pi }{4}\lt x\lt 2\pi \\ \textrm{c}.&\displaystyle \frac{3\pi }{4}\lt x\lt \displaystyle \frac{7\pi }{4} \\ \textrm{d}.&0\lt x\lt \displaystyle \frac{3\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{7\pi }{4}\lt x\lt 2\pi \\ \textrm{e}.&0\lt x\lt \displaystyle \frac{\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{3\pi }{4}\lt x\lt 2\pi \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll}&\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Diketahui}\: \: f(x)=\sin x-\cos x\\ &\textrm{Fungsi}\: \: f\: \: \textrm{naik, jika}\: \: f'(x)>0\\ &\textrm{Selanjutnya}\\ &f'(x)=\cos x+\cos x=0\\ &\sin x=-\cos x\Leftrightarrow \displaystyle \frac{\sin x}{\cos x}=-1\\ &\Leftrightarrow \tan x=-1\\ &\Leftrightarrow \tan x=\tan \displaystyle \frac{3\pi }{4}\\ &\Leftrightarrow x=\displaystyle \frac{3\pi }{4}\pm k.\pi \\\ &\Leftrightarrow k=0\Rightarrow x=\displaystyle \frac{3\pi }{4}\\ &\Leftrightarrow k=1\Rightarrow x=\displaystyle \frac{3\pi }{4}\pm \pi =\frac{7\pi }{4}\\ &\Leftrightarrow k=2\Rightarrow x=\displaystyle \frac{3\pi }{4}\pm 2\pi =\textrm{tm}\\ &\begin{array}{ccccccccc}\\ &&&&&&&\\ &++&&--&&&++&\\\hline 0&&\displaystyle \frac{3\pi }{4}&&&\displaystyle \frac{7\pi }{4}&&2\pi \end{array}\\ &\textrm{ambil titik uji}\: \: x=\displaystyle \frac{1}{2}\pi \\ &\textrm{untuk}\: \: x=\displaystyle \frac{1}{2}\pi \Rightarrow f'\left ( \displaystyle \frac{1}{2}\pi \right )\\ &\quad =\cos \displaystyle \frac{1}{2}\pi +\sin \displaystyle \frac{1}{2}\pi =0+1=1\: \: (\textrm{positif})\\ &\textrm{untuk}\: \: x=\displaystyle \frac{3}{2}\pi \Rightarrow f'\left ( \displaystyle \frac{3}{2}\pi \right )\\ &\quad =\cos \displaystyle \frac{3}{2}\pi +\sin \displaystyle \frac{3}{2}\pi =0-1=-1\: \: (\textrm{negatif})\\ &\textrm{untuk}\: \: x=\displaystyle \frac{11}{6}\pi \Rightarrow f'\left ( \displaystyle \frac{11}{6}\pi \right )\\ &\quad =\cos \displaystyle \frac{11}{6}\pi +\sin \displaystyle \frac{11}{6}\pi =\displaystyle \frac{1}{2}\sqrt{3}-\frac{1}{2}\: \: (\textrm{positif}) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 48.&\textrm{Fungsi}\: \: f(x)=\sin^{2} x\: \: \textrm{dengan}\\ &0\lt x\lt 2\pi \: \: \textrm{naik pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{\pi }{2}\lt x\lt \pi \: \: \textrm{atau}\: \: \displaystyle \frac{3\pi }{2}\lt x\lt 2\pi \\ \textrm{b}.&\displaystyle \frac{2\pi }{3}\lt x\lt \pi \\ \textrm{c}.&0\lt x\lt \displaystyle \frac{\pi }{2}\: \: \textrm{atau}\: \: \pi \lt x\lt \displaystyle \frac{3\pi }{2} \\ \textrm{d}.&\displaystyle \frac{4\pi }{3}\lt x\lt 2\pi \\ \textrm{e}.&\displaystyle \frac{\pi }{3}\lt x\lt \pi \: \: \textrm{atau}\: \: \displaystyle \frac{4\pi }{3}\lt x\lt 2\pi \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll}&\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\textrm{Diketahui}\: \: f(x)=\sin^{2} x\\ &\textrm{Fungsi}\: \: f\: \: \textrm{naik, jika}\: \: f'(x)>0\\ &\textrm{Selanjutnya}\\ &f'(x)=2\sin x\cos x=\sin 2x=0\\ &\Leftrightarrow \sin 2x=0\\ &\Leftrightarrow \sin 2x=\sin 0\\ &\Leftrightarrow 2x=\pm k.2\pi \: \: \textrm{atau}\: \: 2x=\pi \pm k.2\pi \\ &\Leftrightarrow x=\pm k.\pi \: \: \textrm{atau}\: \: x=\frac{\pi}{2} \pm k.\pi \\ &\Leftrightarrow k=0\Rightarrow x=0\: \: \textrm{atau}\: \: x=\displaystyle \frac{\pi }{2}\\ &\Leftrightarrow k=1\Rightarrow x=\pi \: \: \textrm{atau}\: \: x=\displaystyle \frac{\pi }{2}+ \pi =\frac{3\pi }{2}\\ &\Leftrightarrow k=2\Rightarrow x= 2\pi \: \: \textrm{atau}\: \: x=\displaystyle \frac{\pi }{2}+2\pi =\displaystyle \frac{5}{2}\pi \: (\textrm{tm})\\ &\begin{array}{ccccccccc}\\ &&&&&&&&\\ &++&&--&&++&&--&\\\hline 0&&\displaystyle \frac{\pi }{2}&&\pi &&\displaystyle \frac{3\pi }{2}&&2\pi \end{array} \\ &\textrm{ambil titik uji}\: \: x=\displaystyle \frac{1}{6}\pi \\ &\textrm{untuk}\: \: x=\displaystyle \frac{1}{6}\pi \Rightarrow f'\left ( \displaystyle \frac{1}{6}\pi \right )\\ &\quad =\sin 2\left (\displaystyle \frac{1}{6}\pi \right )=\sin \displaystyle \frac{1}{3}\pi =\frac{1}{2}\: \: (\textrm{positif})\\ &\textrm{untuk}\: \: x=\displaystyle \frac{3}{4}\pi \Rightarrow f\left ( \displaystyle \frac{3}{4}\pi \right )\\ &\quad =\sin 2\left (\displaystyle \frac{3}{4}\pi \right )=-1\: \: (\textrm{negatif}) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 49.&\textrm{Fungsi}\: \: f(x)=\cos ^{2}2x\: \: \textrm{untuk}\\ &0^{\circ}\lt x\lt 360^{\circ}\: \: \textrm{turun pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&45^{\circ}\lt x\lt 90^{\circ}\\ \textrm{b}.&135^{\circ}\lt x\lt 180^{\circ}\\ \textrm{c}.&225^{\circ}\lt x\lt 270^{\circ}\\ \textrm{d}.&270^{\circ}\lt x\lt 300^{\circ}\\ \textrm{e}.&315^{\circ}\lt x\lt 360^{\circ} \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&f(x)=\cos ^{2}2x\\ &\textrm{Fungsi}\: \: f\: \: \textrm{turun, jika}\: \: f'(x)<0\\ &f'(x)=2\cos 2x(-\sin 2x)(2)=-2\sin 4x\\ &\textrm{Selanjutnya}\\ &\Leftrightarrow -2\sin 4x=0\Leftrightarrow \sin 4x=0\Leftrightarrow \sin 4x=\sin 0^{\circ}\\ &\Leftrightarrow \begin{cases} 4x=0^{\circ}+k.360^{\circ}&\Rightarrow x=k.90^{\circ}\\ 4x=180^{\circ}+k.360^{\circ}&\Rightarrow x=45^{\circ}+k.90^{\circ} \end{cases}\\ &\Leftrightarrow k=0\Rightarrow x=0^{\circ}\: \: \textrm{atau}\: \: x=45^{\circ}\\ &\Leftrightarrow k=1\Rightarrow x=90^{\circ}\: \: \textrm{atau}\: \: x=135^{\circ}\\ &\Leftrightarrow k=2\Rightarrow x=180^{\circ}\: \: \textrm{atau}\: \: x=225^{\circ}\\ &\Leftrightarrow k=3\Rightarrow x=270^{\circ}\: \: \textrm{atau}\: \: x=315^{\circ}\\ &\Leftrightarrow k=4\Rightarrow x=360^{\circ}\: \: \textrm{atau}\: \: x=405^{\circ}\: \: (\textrm{tm})\\ &\textrm{Gunakan titik uji pada}\: \: x=30^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'(30^{\circ})=-2\sin 4(30^{\circ})=-\sqrt{3}\: \: (\textrm{negatif})\\ &\textrm{Gunakan titik uji pada}\: \: x=60^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'(60^{\circ})=-2\sin 4(60^{\circ})=\sqrt{3}\: \: (\textrm{positif})\\ &\textrm{Gunakan titik uji pada}\: \: x=120^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'(120^{\circ})=-2\sin 4(120^{\circ})=-\sqrt{3}\: \: (\textrm{negatif})\\ &\textrm{Gunakan titik uji pada}\: \: x=150^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'(150^{\circ})=-2\sin 4(150^{\circ})=\sqrt{3}\: \: (\textrm{positif})\\ &\textrm{dan seterusnya}\: ...\\ &\color{black}\begin{array}{ccccccccccc}\\ &&&&&&&&\\ &--&&++&&--&&++\\\hline 0&&\color{red}\displaystyle 45^{\circ}&&90^{\circ}&&\color{red}135^{\circ}&&180^{\circ}\\ &--&&++&&--&&++\\\hline 180^{\circ}&&\color{red}\displaystyle 225^{\circ}&&270^{\circ}&&\color{red}315^{\circ}&&360^{\circ} \end{array} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 50.&(\textbf{SBMPTN 2015})\\ &\textrm{Fungsi}\: \: f(x)=2\sqrt{\sin ^{2}x+\displaystyle \frac{x\sqrt{3}}{2}}\\ & \textrm{pada}\: \: 0\lt x\lt \pi \: \: \textrm{turun pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{5\pi }{12}\lt x\lt \frac{11\pi }{12}\\ \textrm{b}.&\displaystyle \frac{\pi }{12}\lt x\lt \frac{5\pi }{12}\\ \textrm{c}.&\displaystyle \frac{2\pi }{3}\lt x\lt \frac{5\pi }{6}\\ \textrm{d}.&\displaystyle \frac{3\pi }{4}\lt x\lt \pi \\ \textrm{e}.&\displaystyle \frac{3\pi }{4}\lt x\lt \frac{3\pi }{2} \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\textrm{Diketahui}\: \: f(x)=2\sqrt{\sin ^{2}x+\displaystyle \frac{x\sqrt{3}}{2}}\\ &\textrm{Fungsi}\: \: f\: \: \textrm{turun, jika}\: \: f'(x)<0\\ &f'(x)=\displaystyle \frac{\sin 2x+\displaystyle \frac{1}{2}\sqrt{3}}{\sqrt{\sin^{2} x+\displaystyle \frac{x\sqrt{3}}{2}}}=0\\ &\sin 2x+\displaystyle \frac{1}{2}\sqrt{3}=0\Leftrightarrow \sin 2x=-\displaystyle \frac{1}{2}\sqrt{3}\\ &\Leftrightarrow \sin 2x=\sin \displaystyle \frac{4\pi }{3}\\ &\Leftrightarrow 2x=\displaystyle \frac{4\pi }{3}+ k.2\pi \: \: \textrm{atau}\: \: 2x=\pi -\frac{4\pi }{3}+k.2\pi \\ &\Leftrightarrow x=\displaystyle \frac{2\pi }{3}+ k.\pi \: \: \textrm{atau}\: \: x= -\frac{\pi }{6}+k.\pi \\ &\Leftrightarrow k=0\Rightarrow x=\displaystyle \frac{2\pi }{3}\: \: \textrm{atau}\: \: x=-\displaystyle \frac{\pi }{6}\: \: (\textrm{tm})\\ &\Leftrightarrow k=1\Rightarrow x=\displaystyle \frac{5\pi }{3}\: \: \textrm{atau}\: \: x=\displaystyle \frac{5\pi }{6}\\ &\textrm{Gunakan titik uji pada}\: \: x=\displaystyle \frac{\pi }{2}=90^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'\left ( \displaystyle \frac{\pi }{2} \right )=\displaystyle \frac{\sin 2\left ( \displaystyle \frac{\pi }{2} \right )+\displaystyle \frac{1}{2}\sqrt{3}}{\sqrt{\sin^{2} \left ( \displaystyle \frac{\pi }{2} \right )+\displaystyle \frac{\left ( \displaystyle \frac{\pi }{2} \right )\sqrt{3}}{2}}}=+\\ &\qquad (\textrm{positif})\\ &\textrm{Gunakan titik uji pada}\: \: x=\displaystyle \frac{3\pi }{4}=135^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'\left ( \displaystyle \frac{3\pi }{4} \right )=\displaystyle \frac{\sin 2\left ( \displaystyle \frac{3\pi }{4} \right )+\displaystyle \frac{1}{2}\sqrt{3}}{\sqrt{\sin^{2} \left ( \displaystyle \frac{3\pi }{4} \right )+\displaystyle \frac{\left ( \displaystyle \frac{3\pi }{4} \right )\sqrt{3}}{2}}}=-\\ &\qquad (\textrm{negatif})\\ &\begin{array}{cccccc}\\ &&&&\\ &++&&--\\\hline 0&&\displaystyle \displaystyle \frac{2\pi }{3}&&\displaystyle \frac{5\pi }{6} \end{array} \end{aligned} \end{array}$

CONTOH SOAL 9 TURUNAN FUNGSI

 

$\begin{array}{ll}\\ 41.&\textrm{Persamaan garis singgung pada kurva}\\ &y=3\sin x\: \: \textrm{pada titik yang berabsis}\: \: \displaystyle \frac{\pi }{3}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=\displaystyle \frac{2}{3}\left ( x-\frac{\pi }{3} \right )-\frac{2\sqrt{2}}{3}\\ \textrm{b}.&y=\displaystyle \frac{2}{3}\left ( x-\frac{\pi }{3} \right )+\frac{2\sqrt{2}}{3}\\ \textrm{c}.&y=\displaystyle \frac{3}{2}\left ( x-\frac{\pi }{3} \right )-\frac{3\sqrt{3}}{2}\\ \textrm{d}.&y=\displaystyle \frac{3}{2}\left ( x-\frac{\pi }{3} \right )+\frac{3\sqrt{3}}{2}\\ \textrm{e}.&y=\displaystyle \frac{3}{2}\left ( x-\frac{\pi }{3} \right )-\frac{3\sqrt{2}}{2} \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&y=3\sin x,\: \: \textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{3}\\ &y_{0}=3\sin \left ( \displaystyle \frac{\pi }{3} \right )=3\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )=\frac{3\sqrt{3}}{2}\\ &\textrm{kita cari gradien}\: \: m\: \: \textrm{saat}\: \: y',\: \: \textrm{yaitu}:\\ &m=y'=3\cos x,\: \: \textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{3}\\ &m=3\cos \left ( \displaystyle \frac{\pi }{3} \right )=3\left ( \displaystyle \frac{1}{2} \right )=\frac{3}{2}\\ &\textrm{Persamaan garis singgungnya adalah}:\\ &y=m(x-x_{0})+y_{0}\\ &y=\displaystyle \frac{3}{2}\left ( x-\displaystyle \frac{\pi }{3} \right )+\displaystyle \frac{3\sqrt{3}}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 42.&\textrm{Kurva}\: \: y=\sin x+\cos x\: \: \textrm{untuk}\\ &0\lt x\lt \pi \: \: \textrm{memotong sumbu X}\\ &\textrm{di titik A. Persamaan garis}\\ &\textrm{singgung di titik A adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=-\sqrt{2}\left ( x-\displaystyle \frac{\pi }{4} \right )\\ \textrm{b}.&y=-\sqrt{2}\left ( x-\displaystyle \frac{\pi }{2} \right )\\ \textrm{c}.&y=-\sqrt{2}\left ( x-\displaystyle \frac{3\pi }{4} \right )\\ \textrm{d}.&y=\sqrt{2}\left ( x-\displaystyle \frac{\pi }{4} \right )\\ \textrm{e}.&y=\sqrt{2}\left ( x-\displaystyle \frac{3\pi }{4} \right ) \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\textrm{Kurva memotong sumbu X}\\ &\textrm{di titik A, berarti}\: \: y=0\\ &\sin x+\cos x=0\\ &\sin x=-\cos x\\ &\displaystyle \frac{\sin x}{\cos x}=-1\Leftrightarrow \tan x=-1\\ &\tan x=\tan \left ( \displaystyle \frac{3\pi }{4} \right )\Rightarrow x=\displaystyle \frac{3\pi }{4}\\ &\textrm{Jadi, titik A-nya}:\: \: \left ( \displaystyle \frac{3\pi }{4},0 \right )\\ &\textrm{dan nilai gradien}\: \: m=y',\: \: \textrm{yaitu}:\\ &m=\cos x-\sin x\\ &m=\cos \left ( \displaystyle \frac{3\pi }{4} \right )-\sin \left ( \displaystyle \frac{3\pi }{4} \right )\\ &m=-\displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}=-\sqrt{2}\\ &\textrm{Persamaan garis singgung di A}:\\ &y=m(x-x_{0})+y_{0}\\ &\Leftrightarrow \: y=-\sqrt{2}\left ( x-\displaystyle \frac{3\pi }{4} \right ) +0\\ &\Leftrightarrow \: y=-\sqrt{2}\left ( x-\displaystyle \frac{3\pi }{4} \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 43.&\textrm{Persamaan garis singgung pada}\\ &\textrm{kurva}\: \: y=\sec ^{2}x\: \: \textrm{pada titik yang}\\ &\textrm{berabsis}\: \: \displaystyle \frac{\pi }{3}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll} \textrm{a}.&y=8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )-4\\ \textrm{b}.&y=8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )+4\\ \textrm{c}.&y=-8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )-4\\ \textrm{d}.&y=-8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )+4\\ \textrm{e}.&y=4\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )-4 \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&y=\sec^{2} x,\: \: \textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{3}\\ &y_{0}=\sec^{2} \left ( \displaystyle \frac{\pi }{3} \right )=(2)^{2}=4\\ &\textrm{kita cari gradien}\: \: m\: \: \textrm{saat}\: \: y',\: \: \textrm{yaitu}:\\ &m=y'=2\sec^{2} x\tan x,\: \: \textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{3}\\ &m=2\sec^{2} \left ( \displaystyle \frac{\pi }{3} \right )\tan \left ( \displaystyle \frac{\pi }{3} \right )\\ &\quad=2(4)\sqrt{3}=8\sqrt{3}\\ &\textrm{Persamaan garis singgungnya adalah}:\\ &y=m(x-x_{0})+y_{0}\\ &y=8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )+4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 44.&\textrm{Kurva berikut yang memiliki}\\ &\textrm{garis singgung dengan gradien}\\ &4\sqrt{3}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=2\sin x\: \: \textrm{pada titik}\: \: \left ( \displaystyle \frac{\pi }{3},\sqrt{3} \right )\\ \textrm{b}.&y=\cos 2x\: \: \textrm{pada titik}\: \: \left ( \displaystyle \frac{\pi }{12},\frac{1}{2} \right )\\ \textrm{c}.&y=\tan x\: \: \textrm{pada titik}\: \: \left ( \displaystyle \pi ,0 \right )\\ \textrm{d}.&y=2\sec x\: \: \textrm{pada titik}\: \: \left ( \displaystyle \frac{\pi }{3},2 \right )\\ \textrm{e}.&y=\cot x\: \: \textrm{pada titik}\: \: \left ( \displaystyle \frac{\pi }{4},1 \right ) \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{array}{|c|l|l|}\hline \textrm{a}&y=2\sin x&m=2\cos \displaystyle \frac{\pi }{3}\\ &y'=2\cos x&m=2.\displaystyle \frac{1}{2}=1\\\hline \textrm{b}&y=\cos 2x&m=-2\sin 2 \left (\displaystyle \frac{\pi }{12} \right )\\ &y'=-2\sin 2x&m=-2.\displaystyle \frac{1}{2}=-1\\\hline \textrm{c}&y=\tan x&m=\sec^{2} \left (\pi \right )\\ &y'=\sec^{2} x&m=(-1)^{2}=1\\\hline \textrm{d}&y=2\sec x&m=2\sec \left ( \displaystyle \frac{\pi }{3} \right )\tan \left (\displaystyle \frac{\pi }{3} \right )\\ &y'=2\sec x\tan x&m=2.2.\sqrt{3}=4\sqrt{3}\\\hline \textrm{e}&y=\cot x&m=-\csc^{2} \left ( \displaystyle \frac{\pi }{4} \right )\\ &y'=-\csc^{2} x&m=-\left ( \sqrt{2} \right )^{2}=-2\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 45.&\textrm{Persamaan garis singgung pada}\\ &\textrm{kurva}\: \: y=\sec x\: \: \textrm{di titik yang}\\ &\textrm{berabsis}\: \: \displaystyle \frac{\pi }{4}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=\sqrt{3}x-\displaystyle \frac{\sqrt{3}\pi }{4}+\sqrt{3}\\ \textrm{b}.&y=\sqrt{3}x+\displaystyle \frac{\sqrt{3}\pi }{4}+\sqrt{3}\\ \textrm{c}.&y=\sqrt{2}x-\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}\\ \textrm{d}.&y=\sqrt{2}x+\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}\\ \textrm{e}.&y=\sqrt{2}x-\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{3} \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&y=\sec x,\: \: \textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{4}\\ &y_{0}=\sec \left ( \displaystyle \frac{\pi }{4} \right )=\sqrt{2}\\ &\textrm{kita cari gradien}\: \: m\: \: \textrm{saat}\: \: y',\: \: \textrm{yaitu}:\\ &m=y'=\sec x\tan x,\: \: \textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{4}\\ &m=\sec \left ( \displaystyle \frac{\pi }{4} \right )\tan \left ( \displaystyle \frac{\pi }{4} \right )=\sqrt{2}.1=\sqrt{2}\\ &\textrm{Persamaan garis singgungnya adalah}:\\ &y=m(x-x_{0})+y_{0}\\ &\Leftrightarrow \: y=\sqrt{2}\left ( x-\displaystyle \frac{\pi }{4} \right )+\sqrt{2}\\ &\Leftrightarrow \: y=\sqrt{2}x-\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2} \end{aligned} \end{array}$

VISUALISASI KONSEP LIMIT DAN TURUNAN FUNGSI ALJABAR

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Visualisasi Limit & Turunan Fungsi Aljabar

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Pilih Konsep Pembelajaran

Pilih Fungsi Aljabar \(f(x)\)

Persamaan Limit Interaktif

\(f'(x) =\)

lim

h \(\to\)

\(\frac{f(x+h) - f(x)}{h}\)

Ubah nilai \(h\): 1.00

Grafik Sekan & Garis Singgung

Kurva Sekan Tangen

* Geser slider agar nilai \(h \to 0\) untuk melihat garis sekan menempel menjadi garis singgung

Titik Acuan (\(x_0\)): 1.0

Gradien Sekan (\(m\)): --

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VISUALISASI TURUNAN FUNGSI TANGEN

Visualisasi Grafik Turunan Tangen

Hubungan Geometris $f(x) = \tan(x)$ dan $f'(x) = \sec^2(x)$

$$\frac{d}{dx}\tan(x) = \sec^2(x) = 1 + \tan^2(x)$$

Parameter Evaluasi x₀ = 0.00

Nilai Fungsi $f(x_0)$ 1.000 $\tan(x_0)$

Gradien $f'(x_0)$ 2.000 $\sec^2(x_0)$

Segitiga Gradien: m = dy/dx = 2.000

Kemiringan garis singgung oranye bernilai 2.000. Artinya, setiap perubahan kecil $dx$, nilai $y$ melonjak sebesar $2.000 \times dx$.

Kontrol Interaktif

Posisi Titik ($x_0$):

0.79 rad 45.0°

-1.38 rad (-79°) 0.0 rad +1.38 rad (+79°)

Fitur Tampilan: Segitiga Kemiringan ($dy/dx$) Jejak Lintasan $f'(x)$ Tampilkan Asimtot $\pm \frac{\pi}{2}$

Petunjuk Interaktif

• **Geser Grafik Direct:** Anda bisa langsung mengklik atau menggeser (*drag*) area grafik kanan untuk memindahkan titik $x_0$.
• **Laju Pertumbuhan:** Perhatikan bahwa saat $x \to \frac{\pi}{2}$, grafik $\tan(x)$ menjadi sangat curam sehingga nilai $f'(x) = \sec^2(x)$ melambung tinggi secara eksponensial.

Grafik Komparasi Vertikal

$f(x) = \tan(x)$ Garis Singgung $f'(x) = \sec^2(x)$

Grafik Utama: f(x) = tan(x)

Sinkronisasi Domain (x)

Grafik Turunan: f'(x) = sec²(x)

-π/2 (-1.57) Domain Sumbu-X (Radian) +π/2 (+1.57)

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