Monotoniczność funkcji

Wskazówka

<- interaktywna sesja notebooka

from random import choice

from IPython.display import display, Image, Markdown, Latex

import generator_zadan.generatory as gz

print(gz.__version__)

0.2.10

ile_zadan_przykladowych = 10

zadanie = gz.monotonicznosc(choice([1,2,3]))
zadanie

('Zbadać monotoniczność i ekstrema funkcji\n\t\\[\n\t\tf(x)=\\frac{x^{2} + 2 x + 1}{2 - 2 x} \n\t\\]\n',
 "$f'(x)=- \\frac{\\left(x - 3\\right) \\left(x + 1\\right)}{2 \\left(x - 1\\right)^{2}}$\\newline\n$f'(x) > 0 \\textnormal{ dla: }x > -1 \\wedge x < 3 \\wedge x \\neq 1 \\implies f(x) \\textnormal{ rośnie dla } x \\in \\left(-1,1  \\right)\\textnormal{ oraz } x \\in \\left(1,3  \\right) $\\newline\n$f'(x) < 0 \\textnormal{ dla: }3 < x \\vee x < -1 \\implies f(x) \\textnormal{ maleje dla } x \\in \\left( -\\infty, -1  \\right)\\textnormal{ oraz } x \\in \\left(3, \\infty  \\right) $\\newline\n$f'(x) = 0 \\textnormal{ dla } x \\in \\left\\{ -1, \\  3\\right\\} \\implies f_{\\min}\\left(-1\\right) = 0 \\textnormal{ oraz } f_{\\max}\\left(3\\right) = -4$")

Show code cell source Hide code cell source

print("\033[34m** Zadanie **" + '*' * 81 + '\033[0m')
print(zadanie[0])
print("\033[34m*\033[0m" * 95)
print("\033[32m** Rozwiązanie **" + '*' * 78 + '\033[0m')
print(zadanie[1])
print("\033[32m*\033[0m" * 95)

** Zadanie ***********************************************************************************
Zbadać monotoniczność i ekstrema funkcji
	\[
		f(x)=\frac{x^{2} + 2 x + 1}{2 - 2 x} 
	\]

***********************************************************************************************
** Rozwiązanie ********************************************************************************
$f'(x)=- \frac{\left(x - 3\right) \left(x + 1\right)}{2 \left(x - 1\right)^{2}}$\newline
$f'(x) > 0 \textnormal{ dla: }x > -1 \wedge x < 3 \wedge x \neq 1 \implies f(x) \textnormal{ rośnie dla } x \in \left(-1,1  \right)\textnormal{ oraz } x \in \left(1,3  \right) $\newline
$f'(x) < 0 \textnormal{ dla: }3 < x \vee x < -1 \implies f(x) \textnormal{ maleje dla } x \in \left( -\infty, -1  \right)\textnormal{ oraz } x \in \left(3, \infty  \right) $\newline
$f'(x) = 0 \textnormal{ dla } x \in \left\{ -1, \  3\right\} \implies f_{\min}\left(-1\right) = 0 \textnormal{ oraz } f_{\max}\left(3\right) = -4$
***********************************************************************************************

Show code cell source Hide code cell source

for i in range(1, ile_zadan_przykladowych + 1):
    zadanie = gz.monotonicznosc(choice([1,2,3]))
    print(f"\033[34m** Zadanie {i} **" + '*' * 80 + '\033[0m')
    display(Markdown(zadanie[0].split('\n\t')[0]))
    display(Latex('$$' + (zadanie[0].split('\n\t')[2]) + '$$'))
    print("\033[34m*\033[0m" * 95)
    print("\033[32m** Rozwiązanie **" + '*' * 78 + '\033[0m')
    display(Markdown('$$' + zadanie[1].split('$')[1] + '$$'))
    display(Markdown('$$' + zadanie[1].split('$')[3].split('\\implies')[0] + '\\implies' + '$$' ))
    display(Markdown('$$' + '\\implies' + zadanie[1].split('$')[3].split('\\implies')[1] + '$$' ))
    display(Markdown('$$' + zadanie[1].split('$')[5].split('\\implies')[0] + '\\implies' + '$$' ))
    display(Markdown('$$' + '\\implies' + zadanie[1].split('$')[5].split('\\implies')[1] + '$$' ))
    display(Markdown('$$' + zadanie[1].split('$')[7].replace('\\left\\{', '\\left\\lbrace').replace('\\right\\}', '\\right\\rbrace') + '$$'))
    print("\033[32m*\033[0m" * 95)

** Zadanie 1 **********************************************************************************

Zbadać monotoniczność i ekstrema funkcji

\[ f(x)=\frac{x - 1}{x^{2} + 4 x + 4} \]

***********************************************************************************************
** Rozwiązanie ********************************************************************************

$$f’(x)=- \frac{x - 4}{\left(x + 2\right)^{3}}$$

$$f’(x) > 0 \textnormal{ dla: }-2 < x \wedge x < 4 \implies$$

$$\implies f(x) \textnormal{ rośnie dla } x \in \left( -2, 4 \right)$$

$$f’(x) < 0 \textnormal{ dla: }4 < x \vee x < -2 \implies$$

$$\implies f(x) \textnormal{ maleje dla } x \in \left( -\infty, -2 \right)\textnormal{ oraz } x \in \left(4, \infty \right) $$

$$f’(x) = 0 \textnormal{ dla } x \in \left\lbrace 4\right\rbrace \implies f_{\max}\left(4\right) = \frac{1}{12}$$

***********************************************************************************************

** Zadanie 2 **********************************************************************************

Zbadać monotoniczność i ekstrema funkcji

\[ f(x)=\frac{3 x^{2} + 2 x + 2}{4 x + 1} \]

***********************************************************************************************
** Rozwiązanie ********************************************************************************

$$f’(x)=\frac{6 \left(x + 1\right) \left(2 x - 1\right)}{\left(4 x + 1\right)^{2}}$$

$$f’(x) > 0 \textnormal{ dla: }\frac{1}{2} < x \vee x < -1\implies$$

$$\implies f(x) \textnormal{ rośnie dla } x \in \left( -\infty, -1 \right)\textnormal{ oraz } x \in \left(\frac{1}{2}, \infty \right)$$

$$f’(x) < 0 \textnormal{ dla: }x > -1 \wedge x < \frac{1}{2} \wedge x \neq - \frac{1}{4} \implies$$

$$\implies f(x) \textnormal{ rośnie dla } x \in \left(-1,- \frac{1}{4} \right)\textnormal{ oraz } x \in \left(- \frac{1}{4},\frac{1}{2} \right)$$

$$f’(x) = 0 \textnormal{ dla } x \in \left\lbrace -1, \ \frac{1}{2}\right\rbrace \implies f_{\max}\left(-1\right) = -1 \textnormal{ oraz } f_{\min}\left(\frac{1}{2}\right) = \frac{5}{4}$$

***********************************************************************************************

** Zadanie 3 **********************************************************************************

Zbadać monotoniczność i ekstrema funkcji

\[ f(x)=\frac{x^{3} + 3 x^{2} + x - 3}{x + 2} \]

***********************************************************************************************
** Rozwiązanie ********************************************************************************

$$f’(x)=\frac{\left(x + 1\right)^{2} \left(2 x + 5\right)}{\left(x + 2\right)^{2}}$$

$$f’(x) > 0 \textnormal{ dla: }\left(- \frac{5}{2} < x \wedge x < -2\right) \vee \left(-2 < x \wedge x < -1\right) \vee -1 < x \implies$$

$$\implies f(x) \textnormal{ rośnie dla } x \in \left(- \frac{5}{2},-2\right) \textnormal{ oraz } x \in \left(-2,\infty\right) $$

$$f’(x) < 0 \textnormal{ dla: }x < - \frac{5}{2} \implies$$

$$\implies f(x) \textnormal{ maleje dla } x \in \left(-\infty,- \frac{5}{2}\right) $$

$$f’(x) = 0 \textnormal{ dla } x \in \left\lbrace - \frac{5}{2}, \ -1\right\rbrace \implies f_{\min}\left(- \frac{5}{2}\right) = \frac{19}{4}$$

***********************************************************************************************

** Zadanie 4 **********************************************************************************

Zbadać monotoniczność i ekstrema funkcji

\[ f(x)=\frac{3 x^{3} + x + 2}{x + 2} \]

***********************************************************************************************
** Rozwiązanie ********************************************************************************

$$f’(x)=\frac{6 x^{2} \left(x + 3\right)}{\left(x + 2\right)^{2}}$$

$$f’(x) > 0 \textnormal{ dla: }\left(-3 < x \wedge x < -2\right) \vee \left(-2 < x \wedge x < 0\right) \vee 0 < x \implies$$

$$\implies f(x) \textnormal{ rośnie dla } x \in \left(-3,-2\right) \textnormal{ oraz } x \in \left(-2,\infty\right) $$

$$f’(x) < 0 \textnormal{ dla: }x < -3 \implies$$

$$\implies f(x) \textnormal{ maleje dla } x \in \left(-\infty,-3\right) $$

$$f’(x) = 0 \textnormal{ dla } x \in \left\lbrace -3, \ 0\right\rbrace \implies f_{\min}\left(-3\right) = 82$$

***********************************************************************************************

** Zadanie 5 **********************************************************************************

Zbadać monotoniczność i ekstrema funkcji

\[ f(x)=\frac{x + 3}{x^{2} + x - 2} \]

***********************************************************************************************
** Rozwiązanie ********************************************************************************

$$f’(x)=- \frac{\left(x + 1\right) \left(x + 5\right)}{\left(x - 1\right)^{2} \left(x + 2\right)^{2}}$$

$$f’(x) > 0 \textnormal{ dla: }x > -5 \wedge x < -1 \wedge x \neq -2 \implies$$

$$\implies f(x) \textnormal{ rośnie dla } x \in \left(-5,-2 \right)\textnormal{ oraz } x \in \left(-2,-1 \right) $$

$$f’(x) < 0 \textnormal{ dla: }\left(-1 < x \wedge x < 1\right) \vee 1 < x \vee x < -5 \implies$$

$$\implies f(x) \textnormal{ maleje dla } x \in \left( -\infty, -5 \right)\textnormal{ oraz } x \in \left(-1, 1 \right) \textnormal{ oraz } x \in \left(1, \infty \right) $$

$$f’(x) = 0 \textnormal{ dla } x \in \left\lbrace -5, \ -1\right\rbrace \implies f_{\min}\left(-5\right) = - \frac{1}{9} \textnormal{ oraz } f_{\max}\left(-1\right) = -1$$

***********************************************************************************************

** Zadanie 6 **********************************************************************************

Zbadać monotoniczność i ekstrema funkcji

\[ f(x)=\frac{- x^{3} - 3 x^{2} - 3 x - 1}{3 x - 1} \]

***********************************************************************************************
** Rozwiązanie ********************************************************************************

$$f’(x)=- \frac{6 \left(x - 1\right) \left(x + 1\right)^{2}}{\left(3 x - 1\right)^{2}}$$

$$f’(x) > 0 \textnormal{ dla: }\left(-1 < x \wedge x < \frac{1}{3}\right) \vee \left(\frac{1}{3} < x \wedge x < 1\right) \vee x < -1 \implies$$

$$\implies f(x) \textnormal{ rośnie dla } x \in \left(-\infty,\frac{1}{3}\right) \textnormal{ oraz } x \in \left(\frac{1}{3},1\right) $$

$$f’(x) < 0 \textnormal{ dla: }1 < x \implies$$

$$\implies f(x) \textnormal{ maleje dla } x \in \left(1,\infty\right) $$

$$f’(x) = 0 \textnormal{ dla } x \in \left\lbrace -1, \ 1\right\rbrace \implies f_{\max}\left(1\right) = -4$$

***********************************************************************************************

** Zadanie 7 **********************************************************************************

Zbadać monotoniczność i ekstrema funkcji

\[ f(x)=\frac{2 x + 4}{- 2 x^{2} + 2 x + 4} \]

***********************************************************************************************
** Rozwiązanie ********************************************************************************

$$f’(x)=\frac{x \left(x + 4\right)}{\left(x - 2\right)^{2} \left(x + 1\right)^{2}}$$

$$f’(x) > 0 \textnormal{ dla: }\left(0 < x \wedge x < 2\right) \vee 2 < x \vee x < -4 \implies$$

$$\implies f(x) \textnormal{ rośnie dla } x \in \left( -\infty, -4 \right)\textnormal{ oraz } x \in \left(0, 2 \right) \textnormal{ oraz } x \in \left(2, \infty \right) $$

$$f’(x) < 0 \textnormal{ dla: }x > -4 \wedge x < 0 \wedge x \neq -1 \implies$$

$$\implies f(x) \textnormal{ maleje dla } x \in \left(-4,-1 \right)\textnormal{ oraz } x \in \left(-1,0 \right) $$

$$f’(x) = 0 \textnormal{ dla } x \in \left\lbrace -4, \ 0\right\rbrace \implies f_{\max}\left(-4\right) = \frac{1}{9} \textnormal{ oraz } f_{\min}\left(0\right) = 1$$

***********************************************************************************************

** Zadanie 8 **********************************************************************************

Zbadać monotoniczność i ekstrema funkcji

\[ f(x)=\frac{2 x^{2} + x + 1}{1 - 2 x} \]

***********************************************************************************************
** Rozwiązanie ********************************************************************************

$$f’(x)=- \frac{\left(2 x - 3\right) \left(2 x + 1\right)}{\left(2 x - 1\right)^{2}}$$

$$f’(x) > 0 \textnormal{ dla: }x > - \frac{1}{2} \wedge x < \frac{3}{2} \wedge x \neq \frac{1}{2} \implies$$

$$\implies f(x) \textnormal{ rośnie dla } x \in \left(- \frac{1}{2},\frac{1}{2} \right)\textnormal{ oraz } x \in \left(\frac{1}{2},\frac{3}{2} \right) $$

$$f’(x) < 0 \textnormal{ dla: }\frac{3}{2} < x \vee x < - \frac{1}{2} \implies$$

$$\implies f(x) \textnormal{ maleje dla } x \in \left( -\infty, - \frac{1}{2} \right)\textnormal{ oraz } x \in \left(\frac{3}{2}, \infty \right) $$

$$f’(x) = 0 \textnormal{ dla } x \in \left\lbrace - \frac{1}{2}, \ \frac{3}{2}\right\rbrace \implies f_{\min}\left(- \frac{1}{2}\right) = \frac{1}{2} \textnormal{ oraz } f_{\max}\left(\frac{3}{2}\right) = - \frac{7}{2}$$

***********************************************************************************************
** Zadanie 9 **********************************************************************************

Zbadać monotoniczność i ekstrema funkcji

\[ f(x)=\frac{- x^{2} + 2 x - 1}{2 x - 1} \]

***********************************************************************************************
** Rozwiązanie ********************************************************************************

$$f’(x)=- \frac{2 x \left(x - 1\right)}{\left(2 x - 1\right)^{2}}$$

$$f’(x) > 0 \textnormal{ dla: }x > 0 \wedge x < 1 \wedge x \neq \frac{1}{2} \implies$$

$$\implies f(x) \textnormal{ rośnie dla } x \in \left(0,\frac{1}{2} \right)\textnormal{ oraz } x \in \left(\frac{1}{2},1 \right) $$

$$f’(x) < 0 \textnormal{ dla: }1 < x \vee x < 0 \implies$$

$$\implies f(x) \textnormal{ maleje dla } x \in \left( -\infty, 0 \right)\textnormal{ oraz } x \in \left(1, \infty \right) $$

$$f’(x) = 0 \textnormal{ dla } x \in \left\lbrace 0, \ 1\right\rbrace \implies f_{\min}\left(0\right) = 1 \textnormal{ oraz } f_{\max}\left(1\right) = 0$$

***********************************************************************************************

** Zadanie 10 **********************************************************************************

Zbadać monotoniczność i ekstrema funkcji

\[ f(x)=\frac{3 x^{3} - x - 1}{- 2 x - 2} \]

***********************************************************************************************
** Rozwiązanie ********************************************************************************

$$f’(x)=- \frac{3 x^{2} \left(2 x + 3\right)}{2 \left(x + 1\right)^{2}}$$

$$f’(x) > 0 \textnormal{ dla: }x < - \frac{3}{2} \implies$$

$$\implies f(x) \textnormal{ rośnie dla } x \in \left(-\infty,- \frac{3}{2}\right) $$

$$f’(x) < 0 \textnormal{ dla: }\left(- \frac{3}{2} < x \wedge x < -1\right) \vee \left(-1 < x \wedge x < 0\right) \vee 0 < x \implies$$

$$\implies f(x) \textnormal{ maleje dla } x \in \left(- \frac{3}{2},-1\right) \textnormal{ oraz } x \in \left(-1,\infty\right) $$

$$f’(x) = 0 \textnormal{ dla } x \in \left\lbrace - \frac{3}{2}, \ 0\right\rbrace \implies f_{\max}\left(- \frac{3}{2}\right) = - \frac{77}{8}$$

***********************************************************************************************