Contoh Soal 3 Turunan Fungsi Aljabar

 $\begin{array}{ll}\\ 11.&\textrm{Jika}\: \: \: f(x)=\displaystyle (x^{2}+2)\sqrt{x^{2}+x+3}\\&\textrm{maka}\: \: {f}\, '(2)=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 18 &&\textrm{d}.\quad 13 \\\\ \textrm{b}.\quad \displaystyle \color{red}17 \quad &\textrm{c}.\quad \displaystyle 15 \quad &\textrm{e}.\quad 12 \end{array}\\\\ &\textbf{Jawab}:\\&\begin{aligned}f(x)&=\displaystyle (x^{2}+2)\sqrt{x^{2}+x+3} \\ {f}\, '(x)&=(2x)\sqrt{x^{2}+x+3}\\ &+(x^{2}+2).\displaystyle \frac{1}{2}\left ( x^{2}+x+3 \right )^{^{-\frac{1}{2}}}.\left ( 2x+1 \right ) \\ &=2x\sqrt{x^{2}+x+3}+\displaystyle \frac{(x^{2}+2)(2x+1)}{2\sqrt{x^{2}+x+3}} \\ {f}\, '(2)&=2(2)\sqrt{(2)^{2}+(2)+3}+\displaystyle \frac{((2)^{2}+2)(2(2)+1)}{2\sqrt{(2)^{2}+(2)+3}} \\ &=4\sqrt{9}+\displaystyle \frac{6.5}{2\sqrt{9}} \\ &=4.3+\displaystyle \frac{6.5}{2.3}\\ &=12+5\\ &=17 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Jika rusuk suatu kubus bertambah panjang }\\ &\textrm{dengan laju 7}\: \: cm/detik\: , \: \textrm{maka laju}\\ &\textrm{bertambahnya volume pada saat rusuk }\\ &\textrm{panjangnya 15}\: \: cm\: \: \textrm{adalah}.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 675 \: \: \: \: cm^{3}/detik &\textrm{d}.\quad \color{red}4725 \: \: \: \: cm^{3}/detik \\ \textrm{b}.\quad \displaystyle 1575 \: \: \: \: cm^{3}/detik \quad  &\textrm{e}.\quad 23625 \: \: \: \: cm^{3}/detik\\ \textrm{c}.\quad \displaystyle 3375 \: \: \: \: cm^{3}/detik \quad \end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}\textrm{Laju }&\textrm{pertambahan volumenya}:\\ &\begin{cases} \bullet \: \: \: \frac{\mathrm{d} s}{\mathrm{d} t} & = 7\: \: cm/detik\\ \bullet \: \: \: V & =s^{3} \rightarrow \textrm{d}V=3s^{2}\: \: \textrm{d}s\: \: \: \textrm{atau}\\ \, \: \: \: \: \frac{\mathrm{d} V}{\mathrm{d} s} & =3s^{2}\: \: cm^{3}/cm\rightarrow s=15\: \: cm \end{cases}\\ \frac{\mathrm{d} V}{\mathrm{d} t}&=\frac{\mathrm{d} V}{\mathrm{d} t}\\ &=\frac{\mathrm{d} V}{\mathrm{d} s}\times \frac{\mathrm{d} s}{\mathrm{d} t}\\ &=3s^{2} \: \: \: \: cm^{3}/cm \times 7\: \: \: \: cm/detik \\ &=3(15)^{2}\times 7\: \: \: \: cm^{3}/detik\\ &=4725\: \: \: \: cm^{3}/detik \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{persamaan garis singgung di}\: \: x=1\\ &\textrm{pada kurva}\: \: y=x^{3}-3x^{2}+1\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}y=-3x+2&\textrm{d}.\quad y=3x-2\\ \textrm{b}.\quad y=-3x+4\quad  &\textrm{e}.\quad y=-3x+3\\ \textrm{c}.\quad y=3x-4\quad \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{array}{|l|}\hline \textrm{Titik singgung di}\: \: x=1\\\hline \begin{aligned} y&=x^{3}-3x^{2}+1\\ y&=(1)^{3}-3(1)^{2}+1\\ &=1-3+1\\ &=-1\\ &\textrm{titik}\: (a,b)=(1,-1) \end{aligned}\\\hline \textrm{Gradien garis singgung di}\: \: x=1\\\hline \begin{aligned} {y}\, '=m\, _{_{x=1}}&=3x^{2}-6x\\ &=3(1)^{2}-6(1)\\ &=3-6\\ &=-3 \end{aligned}\\\hline \textrm{Persamaan garis singgung}\\\hline \begin{aligned}y&=m(x-a)+b\\ &=-3(x-1)+(-1)\\ &=-3x+3-1\\ &=-3x+2\end{aligned}\\\hline \end{array}\\ &\textrm{Berikut ilustrasi gambarnya}  \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Suatu kurva}\: \: y=x^{3}+2ax^{2}+b\\ &\textrm{Sebuah garis}\: \: y=-9x-2\: \: \textrm{menyinggung}\\ &\textrm{kurva di titik dengan} \: \: x=1,\\ &\textrm{maka nilai}\: \: a \: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}-3&&\textrm{d}.\quad 3\\ \textrm{b}.\quad -\displaystyle \frac{1}{3} \quad &\textrm{c}.\quad \displaystyle \frac{1}{3} \quad &\textrm{e}.\quad 8 \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Gradien garis singgung}\\ &y\: _{_{_{di\: \: x=1}}}\begin{cases} y & =x^{3}+2ax^{2}+b \\ &\rightarrow m={y}\, '=\color{blue}3x^{2}+4ax\\\\ y & =-9x-2 \rightarrow m=y'= \color{blue}-9 \end{cases}\\ &\textrm{Sehingga}\:, \\ &m=m={y}\, '\\ &-9=3x^{2}+4ax\\ &-9=3(1)^{2}+4a(1)\\ &-9=3+4a\\ &-4a=3+9\\ &a=\displaystyle \frac{12}{-4}\\ &=-3\\ & \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Jika grafik fungsi}\: \: f(x)=x^{3}+ax^{2}+bx+c\\ &\textrm{hanya turun untuk interval}\: \: -1< x< 5\, ,\\ &\textrm{maka nilai}\: \: a+b \: \: \textrm{adalah}....\\&\begin{array}{lll}\\ \textrm{a}.\quad \color{red}-21&&\textrm{d}.\quad 21\\ \textrm{b}.\quad -\displaystyle 9 \quad &\textrm{c}.\quad \displaystyle 9 \quad &\textrm{e}.\quad 24 \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\\ &f(x)=x^{3}+ax^{2}+bx+c\\ &\textrm{grafik fungsi turun}\: \: \textrm{berarti}:\: \: \: {f}\, '(x)< 0\\ &{f}\, '(x)< 0\\ &\Leftrightarrow 3x^{2}+2ax+b< 0\\ &\Leftrightarrow (x+1)(x-5)< 0 \\ &\textrm{ini maksud pada interval}\\ &-1< x< 5\: \: \: \textrm{pada soal}\\ &x^{2}-4x-5< 0\\ &\textrm{dikalikan dengan 3 supaya terjadi persamaan}\\ &3x^{2}-3.4x-3.5=3x^{2}+2ax+b< 0\\ &3x^{2}+2(-6)x+(-15)=3x^{2}+2ax+b< 0\\ &\begin{cases} a & =-6 \\ b & = -15 \end{cases}\\ &\textrm{Sehingga}\: ,\\ &a+b=-6+(-15)=-21 \end{aligned} \end{array}$.