جواب تمرین صفحه 125 درس 5 حسابان دوازدهم (کاربردهای مشتق)

الف

ب

پ

الف

ب

پ

الف

 \(\begin{array}{l}f'(x) = 6x - 2 \Rightarrow 6x - 2 = 0 \Rightarrow x = \frac{1}{3}\\\left\{ \begin{array}{l}x = - 2 \Rightarrow y = 12 + 4 + 5 = 21\\x = \frac{1}{3} \Rightarrow y = \frac{1}{3} - \frac{2}{3} + 5 = \frac{{14}}{3}\\x = 1 \Rightarrow y = 3 - 2 + 5 = 6\end{array} \right.\end{array}\)

\(21 \to \max \)اکسترمم مطلق

اکسترمم مطلق\(\frac{{14}}{3} \to \min \)

ب

 \(\begin{array}{l}f'(x) = 3{x^2} - 3 \Rightarrow 3{x^2} - 3 = 0 \Rightarrow x = \pm 1\\\left\{ \begin{array}{l}x = - 1 \Rightarrow y = - 1 + 3 = 2\\x = 1 \Rightarrow y = 1 - 3 = - 2\\x = 2 \Rightarrow y = 8 - 6 = 2\end{array} \right.\end{array}\)

اکسترمم مطلق\(2 \to \max \)

\( - 2 \to \min \) اکسترمم مطلق

اکسترمم مطلق\(2 \to \max \)

پ

 \(f'(x) = \left\{ \begin{array}{l}\frac{1}{{2\sqrt x }}\quad \quad 0 < x < 2\\ - 1\quad \quad \quad \;\;x > 2\end{array} \right.\)

\(\begin{array}{l}\left( {1\;,\;2} \right) \Rightarrow 1 + a + b = 2 \Rightarrow a + b = 1\quad \left( i \right)\\f'(x) = 3{x^2} + a \Rightarrow f'(1) = 0 \Rightarrow 3 + a = 0 \Rightarrow a = - 3\\\left( i \right):\quad \Rightarrow - 3 + b = 1 \Rightarrow b = 4\end{array}\)

\(\begin{array}{l}x\;y = 100 \Rightarrow y = \frac{{100}}{x}\\V = 2(x - 4)(y - 4) = 2xy - 8x - 8y + 32\\V(x) = 232 - 8x - \frac{{800}}{x} = \frac{{232x - 8{x^2} - 800}}{x}\\ \Rightarrow V'(x) = \frac{{(232 - 16x)x - 1(232x - 8{x^2} - 800)}}{{{x^2}}}\\ \Rightarrow V'(x) = \frac{{ - 8{x^2} + 800}}{{{x^2}}} = 0 \Rightarrow - 8{x^2} + 800 \Rightarrow {x^2} = 100\\ \Rightarrow x = 10\\ \Rightarrow y = 10\end{array}\)

\(\begin{array}{l}y = \sqrt {16 - {x^2}} \\S = 2x.y = 2x\sqrt {16 - {x^2}} \\S'(x) = 2\sqrt {16 - {x^2}} + \frac{{2x( - 2x)}}{{2\sqrt {16 - {x^2}} }} = \frac{{2(16 - {x^2}) - 2{x^2}}}{{\sqrt {16 - {x^2}} }}\\S'(x) = \frac{{32 - 4{x^2}}}{{\sqrt {16 - {x^2}} }} = 0 \Rightarrow 4{x^2} = 32 \Rightarrow {x^2} = 8 \Rightarrow x = \sqrt 8 \\ \Rightarrow y = \sqrt 8 \end{array}\)

الف

\(\begin{array}{l}f(x) = 2{x^3} - 3{x^2} - 12x + 7\\f'(x) = 6{x^2} - 6x - 12 = 0 \Rightarrow {x^2} - x - 2 = 0 \Rightarrow (x - 2)(x + 1) = 0 \Rightarrow \left\{ \begin{array}{l}x = - 1\\x = 2\end{array} \right.\end{array}\)

در بازه های    و   صعودی و در بازه ی (-1,2) نزولی.

ب

\(\begin{array}{l}f(x) = \frac{x}{{x - 2}}\\f'(x) = \frac{{1(x - 2) - 1(x)}}{{{{(x - 2)}^2}}} = \frac{{ - 2}}{{{{(x - 2)}^2}}} < 0\end{array}\)