جواب تمرین صفحه 144 درس 5 حسابان دوازدهم (کاربردهای مشتق)

الف \(\begin{array}{l}(1):\quad {D_f} = \mathbb{R}\\(2):\quad f(x) = 0 \Rightarrow 2{x^2} - 4x + 1 = 0 \Rightarrow x = 1 \pm \frac{{\sqrt 2 }}{2}\\(3):\quad f'(x) = 4x - 4 = 0 \Rightarrow x = 1\\(4):\quad f''(x) = 4 > 0\\(5):\quad x = 0 \Rightarrow y = 1\end{array}\)

ب \(\begin{array}{l}(1):\quad {D_f} = \mathbb{R}\\\\(2):\quad f'(x) = 3{x^2} - 5 = 0 \Rightarrow \left\{ \begin{array}{l}x = \frac{{\sqrt 5 }}{3} \simeq 1/3 \Rightarrow y = 0/7\\\\x = - \frac{{\sqrt 5 }}{3} \simeq - 1/3 \Rightarrow y = 9/3\end{array} \right.\\\\(3):\quad f''(x) = 6x = 0 \Rightarrow x = 0 \Rightarrow y = 5\end{array}\)

پ \(\begin{array}{l}(1):\quad {D_f} = \mathbb{R}\\\\(2):\quad f(x) = 0 \Rightarrow \left\{ \begin{array}{l}x = 0 \Rightarrow y = 0\\\\x = - 2 \Rightarrow y = 0\end{array} \right.\\(3):\quad f'(x) = - 3{x^2} - 8x - 4 = 0 \Rightarrow \left\{ \begin{array}{l}x = - 4 \Rightarrow y = 16\\\\x = - \frac{1}{2} \Rightarrow y = \frac{9}{8}\end{array} \right.\\\\(3):\quad f''(x) = - 6x - 8 = 0 \Rightarrow x = - \frac{4}{3} \Rightarrow y = \frac{{16}}{{27}}\end{array}\)

ت \(\begin{array}{l}(1):\quad x - 2 = 0 \Rightarrow x = 2 \Rightarrow {D_f} = \mathbb{R} - \left\{ 2 \right\}\\(2):\quad \mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{2x - 1}}{{x - 2}} = \frac{3}{{{0^ - }}} = - \infty \\(3):\quad \mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{2x - 1}}{{x - 2}} = \frac{3}{{{0^ + }}} = \infty \\(4):\quad \mathop {\lim }\limits_{x \to \pm \infty } f(x) = \mathop {\lim }\limits_{x \to \pm \infty } \frac{{2x - 1}}{{x - 2}} = 2\\(5):\quad f'(x) = \frac{{ - 3}}{{{{(x - 2)}^2}}} < 0\end{array}\)

ث \(\begin{array}{l}(1):\quad x + 3 = 0 \Rightarrow x = - 3 \Rightarrow {D_f} = \mathbb{R} - \left\{ { - 3} \right\}\\(2):\quad \mathop {\lim }\limits_{x \to - {3^ - }} f(x) = \mathop {\lim }\limits_{x \to - {3^ - }} \frac{{ - x}}{{x + 3}} = \frac{3}{{{0^ - }}} = - \infty \\(3):\quad \mathop {\lim }\limits_{x \to - {3^ + }} f(x) = \mathop {\lim }\limits_{x \to - {3^ + }} \frac{{ - x}}{{x + 3}} = \frac{3}{{{0^ + }}} = \infty \\(4):\quad \mathop {\lim }\limits_{x \to \pm \infty } f(x) = \mathop {\lim }\limits_{x \to \pm \infty } \frac{{ - x}}{{x + 3}} = - 1\\(5):\quad f'(x) = \frac{{ - 3}}{{{{(x + 3)}^2}}} < 0\end{array}\)

ج \(\begin{array}{l}(1):\quad {D_f} = \mathbb{R}\\\\(3):\quad f'(x) = 6{x^2} - 18x + 12 = 0 \Rightarrow \left\{ \begin{array}{l}x = 1 \Rightarrow y = 6\\\\x = 2 \Rightarrow y = 5\end{array} \right.\\\\(3):\quad f''(x) = 12x - 18 = 0 \Rightarrow x = \frac{3}{2} \Rightarrow y = \frac{{11}}{2}\end{array}\)

\(\begin{array}{l}(2\;,\;1) = ( - \frac{d}{c}\;,\;\frac{a}{c}) \Rightarrow - \frac{d}{c} = 2\;,\;\frac{a}{c} = 1 \Rightarrow \left\{ \begin{array}{l}a = c\\\\d = - 2c\end{array} \right.\\\\( - 1\;,\;0) \Rightarrow a( - 1) + b = 0 \Rightarrow a = b\\\\ \Rightarrow f(x) = \frac{{ax + a}}{{ax - 2a}} \Rightarrow f(x) = \frac{{x + 1}}{{x - 2}}\end{array}\)

\(f'(x) = 3{x^2} + 1 \Rightarrow f''(x) = 6x\)

نقطه عطف \( \Rightarrow 6x = 0 \Rightarrow x = 0 \Rightarrow f(0) = - 2\)

تابع اکیداً صعودی \(f'(x) = 3{x^2} + 1 \Rightarrow f'(x) > 0 \Rightarrow \)

\(a > 0\) : ضریب \({x^3}\)

در نتیجه گزینه (ب) جواب صحیح است.