جواب تمرین صفحه 136 درس 5 حسابان دوازدهم (کاربردهای مشتق)

الف \(f\left( x \right) = \frac{1}{3}{x^3} - {x^2} - 3x + 4 \Rightarrow {D_f} = \mathbb{R}\)

\( \Rightarrow f'(x) = {x^2} - 3 \Rightarrow f''(x) = 2x = 0 \Rightarrow x = 0\)

ب \(f\left( x \right) = \frac{{x + 1}}{{x - 1}} \Rightarrow x - 1 = 0 \Rightarrow x = 1 \Rightarrow {D_f} = \mathbb{R} - \left\{ 1 \right\}\)

\( \Rightarrow f'(x) = \frac{{ - 2}}{{{{(x - 1)}^2}}} \Rightarrow f''(x) = \frac{4}{{{{(x - 1)}^3}}}\)

پ \(f\left( x \right) = \sqrt[3]{{x + 1}} \Rightarrow {D_f} = \mathbb{R}\)

\(\begin{array}{l} \Rightarrow f'(x) = \frac{1}{{3\sqrt[3]{{{{(x + 1)}^2}}}}} \Rightarrow f''(x) = \frac{{ - 2}}{{9\sqrt[3]{{{{(x + 1)}^5}}}}}\\ \Rightarrow x + 1 = 0 \Rightarrow x = - 1\end{array}\)

الف \(y = {x^3}\)

ب \(y = {(x - 1)^3}\)

پ \(y = {x^3} + 1\)

ت \(y = {(x - 2)^3} + 2\)

 \(\begin{array}{l}f(x) = a\;{x^3} + b{x^2} + c0 + 0 + c = 1 \Rightarrow \\a + b + 1 = 2 \Rightarrow a + b = 1\\x = \frac{1}{2} \Rightarrow f''(\frac{1}{2}) = 0 \Rightarrow 6a\;x + 2b \Rightarrow 6a\;(\frac{1}{2}) + 2b = 0 \Rightarrow 3a + 2b = 0\\\left\{ \begin{array}{l}a + b = 1\\3a + 2b = 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a = - 2\\b = 3\end{array} \right.\end{array}\)

\(\begin{array}{l}y' = 3{x^2} + 2ax + b\\y'' = 6x + 2a \Rightarrow 6(0) + 2a = 0 \Rightarrow \\\min :\quad x = 2 \Rightarrow 3{(2)^2} + 0 + b = 0 \Rightarrow \\(0\;,\;0):\quad 0 + 0 + 0 + c = 0 \Rightarrow \end{array}\)