Jawab: D

\(y = \sqrt{x}\)

Untuk x = 4 → y = 2

Untuk x = 3,96 → y = ?

Δx = 0,04 → Δy = ?

\begin{equation*}
\begin{split}
\frac {\Delta y}{\Delta x} & = \frac {dy}{dx} \\\\
\frac {\Delta y}{0,04} & = \frac 12 x^{-\frac 12} \\\\
\frac {\Delta y}{0,04} & = \frac {1}{2\sqrt{x}} \\\\
\frac {\Delta y}{0,04} & = \frac {1}{2\sqrt{4}} \\\\
\frac {\Delta y}{0,04} & = \frac 14 \\\\
\Delta y & = 0,01
\end{split}
\end{equation*}

Maka \(\sqrt{3,96} = 2 - 0,01 = 1,99\)

Jawab: A

Volume kubus = s³ dan luas permukaan kubus =  6s²

Saat volume kubus 64 mm³ → s = 4 mm

Laju perubahan volume

\begin{equation*}
\begin{split}
\frac {dV}{dt} & = \frac {dV}{ds} \:.\: \frac {ds}{dt} \\\\
2,5 & = 3s^2 \:.\: \frac {ds}{dt} \\\\
2,5 & = 3s^2 \:.\: \frac {ds}{dt} \\\\
2,5 & = 3 \:.\: 4^2 \:.\: \frac {ds}{dt} \\\\
2,5 & = 48 \:.\: \frac {ds}{dt} \\\\
\frac {2,5}{48} & = \frac {ds}{dt}
\end{split}
\end{equation*}

Laju perubahan luas

\begin{equation*}
\begin{split}
\frac {dA}{dt} & = \frac {dA}{ds} \:.\: \frac {ds}{dt} \\\\
\frac {dA}{dt} & = 12s \:.\: \frac {ds}{dt} \\\\
\frac {dA}{dt} & = 12 \:.\: 4 \:.\: \frac {2,5}{48} \\\\
\frac {dA}{dt} & = 2,5 \text{ mm}^2/s
\end{split}
\end{equation*}

Jawab: 

Jawab: B

Titik stationer

\begin{equation*}
\begin{split}
f(x) & = x^3 + 3 x^2 - 9x + 3 \\\\
f'(x) & = 3x^2 + 6x - 9 \\\\
f'(x) & = x^2 + 2x - 3 \\\\
0 & = (x + 3)(x - 1) \\\\
x & = -3 \text{ atau } x = 1
\end{split}
\end{equation*}

x = −3 di luar interval, tidak digunakan

x = 1 di dalam interval, akan digunakan

Nilai maksimum dan minimum dalam inverval \(0 \leq x \leq 3\)

\begin{equation*}
\begin{split}
f(x) & = x^3 + 3 x^2 - 9x + 3 \\\\
f(1) & = -2 \quad {\color {red} \text{minimum}} \\\\
f(0) & = 3 \\\\
f(3) & = 30 \quad {\color {red} \text{maksimum}}
\end{split}
\end{equation*}

Jawab: D

Persamaan \(y^2 - x^2 = 4\) dapat diubah menjadi \(y = \sqrt{x^2 + 4}\)

Misalkan titik pada kurva (x,y), maka jarak antara titik (2,0) dan (x,y):

\begin{equation*}
\begin{split}
d & = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\\\
d & = \sqrt{(x - 2)^2 + (y - 0)^2} \\\\
d & = \sqrt{(x - 2)^2 + \left(\sqrt{x^2 + 4}\right)^2} \\\\
d & = \sqrt{x^2 - 4x + 4 + x^2 + 4} \\\\
d & = (2x^2 - 4x + 8)^{\frac 12} \\\\
d' & = \frac 12 (4x - 4) (2x^2 - 4x + 8)^{-\frac 12} \\\\
d' & = \frac {(4x - 4)}{2 \sqrt{2x^2 - 4x + 8}}\\\\
\end{split}
\end{equation*}

Jarak terdekat, d' = 0

\begin{equation*}
\begin{split}
0 & = \frac {(4x - 4)}{2 \sqrt{2x^2 - 4x + 8}}\\\\
0 & = 4x - 4 \\\\
x & = 1
\end{split}
\end{equation*}

Substitusi x = 1 ke persamaan d

\begin{equation*}
\begin{split}
d & = \sqrt{2x^2 - 4x + 8} \\\\
d & = \sqrt{2 \:.\: 1^2 - 4 \:.\: 1 + 8} \\\\
d & = \sqrt{6}
\end{split}
\end{equation*}