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Project: am_PS3_1

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Image: ubuntu2204

Kernel: SageMath 10.1

In [4]:

$\displaystyle \left[x = -5 \, y + 9\right]$

Analiza matematyczna - PS

Informatyka, sem.I, studia niestacjonarne I stopnia, 2023/24

Lista nr 1: Funkcje jednej zmiennej. Własności funkcji

Zad. 1. Sprawdzić, czy podane funkcje są rosnące na wskazanych zbiorach:

a) $f(x)=x^2,\ x\in \langle 0;\infty)$

b) $g(x)=\displaystyle\frac{1}{x^4+1},\ x\in(-\infty;0\rangle$

c) $h(x)=\sqrt[3]{x},\ x\in(-\infty;0\rangle;$

d) $p(x)=\sqrt{x+1},\ x\in\langle -1;\infty)$

In [8]:

#a
x1,x2 = var('x1,x2')
assume(x1<x2, x1>=0)

f(x) = x^2
bool(f(x1)<f(x2))

In [9]:

#b
forget()
x1,x2 = var('x1,x2')
assume(x1<x2, x2<=0)

f(x) = 1/(x^4+1)
bool(f(x1)<f(x2))

True

In [15]:

#c
forget()
x1,x2 = var('x1,x2')
assume(x1<x2, x2<=0)

f(x) = x**(1/3)
show(bool(f(x1)<f(x2)))
show(plot(f(x)))
#sagemath się wykłada

$\displaystyle \mathrm{False}$

verbose 0 (3935: plot.py, generate_plot_points) WARNING: When plotting, failed to evaluate function at 100 points.
verbose 0 (3935: plot.py, generate_plot_points) Last error message: 'Unable to compute f(-0.001334360219118633)'

In [0]:

In [16]:

#d
forget()
x1,x2 = var('x1,x2')
assume(x1<x2, x1>=-1)

f(x) = (x+1)**(1/2)
bool(f(x1)-f(x2)<0)

False

Zad. 2. Sprawdzić, czy podane funkcje są malejące na wskazanych zbiorach:

a) $f(x)=3-4x,\ x\in\mathbb{R}$

b) $g(x)=x^2-2x,\ x\in(-\infty;1\rangle$

c) $h(x)=\displaystyle\frac{1}{1+x^2},\ x\in\langle 0;\infty)$

d) $p(x)=\displaystyle\frac{1}{1+x},\ x\in(-\infty;-1)$

In [0]:

In [17]:

#a
forget()
x1,x2 = var('x1,x2')
assume(x1<x2)

f(x) = 3-4*x
bool(f(x1)>f(x2))

True

In [22]:

#b
reset()
forget()
x1,x2 = var('x1,x2')
assume(x1<x2, x2<=0)

g(x) = x^2-x*2
bool(g(x1)>g(x2))

True

In [14]:

#c
forget()
x1,x2 = var('x1,x2')
assume(x1<x2, x1>0)

h(x) = 1/(1+x^2)
bool(h(x1)-h(x2)>0)

True

In [13]:

#d
forget()
x1,x2 = var('x1,x2')
assume(x1<x2, x1>-1)

p(x) = 1/(1+x)
bool(p(x1)-p(x2)>0)

True

Zad. 3. Sprawdzić, czy podane funkcje są różnowartościowe na wskazanych zbiorach:

a) $f(x)=x^3+1,\ x\in\mathbb{R}$

b) $g(x)=\displaystyle\frac{1}{x^2},\ x\in(-\infty;0)$

c) $h(x)=\sqrt{x}+1,\ x\in\langle 0;\infty)$

In [33]:

#a
forget()
x1,x2 = var('x1,x2')
#assume(f(x1)==f(x2))
f(x) = x^3 + 1
# równoważnoc implikacji a=>b <=> ~a=>~b
show(bool(x1==x2))

#assume(x1==x2)
#show(bool(f(x1)==f(x2)))

show(solve(f(x1)==f(x2),x1))

$\displaystyle \mathrm{False}$

$\displaystyle \left[x_{1} = \frac{1}{2} \, x_{2} {\left(i \, \sqrt{3} - 1\right)}, x_{1} = -\frac{1}{2} \, x_{2} {\left(i \, \sqrt{3} + 1\right)}, x_{1} = x_{2}\right]$

In [34]:

#b
forget()
x1,x2 = var('x1,x2')
assume(x1!=x2, x1<0)

g(x) = 1/(x^2)
bool(g(x1)!=g(x2))

False

In [35]:

#c
forget()
x1,x2 = var('x1,x2')
assume(x1!=x2, x1>=0)

g(x) = x**1/2 + 1
bool(g(x1)!=g(x2))

True

Zad. 4. Sprawdzić, które z podanych funkcji są parzyste, a które nieparzyste:

a) $\displaystyle f(x)=x^4-3x^2+1$

b) $\displaystyle g(x)=2^x+2^{-x}$

c) $\displaystyle h(x)=|\sin x|$

d) $\displaystyle p(x)=\frac{\sin x}{x^3}$

e) $\displaystyle f(x)=\frac{2+x^2}{x^5},$

f) $\displaystyle g(x)=\sin^3 x$

g) $\displaystyle h(x)=3^x-3^{-x}$

h) $\displaystyle p(x)=x|x|$

In [4]:

#a
forget()
x1,x2 = var('x1,x2')
assume(x1==-x2)

g(x) = x^4 -3*x^2 + 1
bool(g(x1)==g(x2))

True

In [3]:

#b
forget()
x1,x2 = var('x1,x2')
assume(x1==-x2)

g(x) = 2^x -2^-x
print(bool(g(x1)==g(x2)))
print(bool(g(x1)==-g(-x2)))

False
False

In [8]:

#c
forget()
x1,x2 = var('x1,x2')
assume(x1==-x2)

g(x) = abs(sin(x))
bool(g(x1)==-g(-x2))

False

In [2]:

#d
forget()
x1,x2 = var('x1,x2')
assume(x1==-x2)

g(x) = sin(x)/(x^3)
bool(g(x1)==-g(-x2))

False

Zad. 5. Określić funkcje złożone $f\circ f$ , $f\circ g$ , $g\circ f$ , $g\circ g$ oraz ich dziedziny, jeżeli:

a) $\displaystyle f(x)=|x|, \ g(x)=-3x+2$

b) $\displaystyle f(x)=\sqrt{x},\ g(x)=x^3+1$

c) $\displaystyle f(x)=x^2,\ g(x)=\sqrt{x},$

d) $\displaystyle f(x)=2^x,\ g(x)=\cos x$

e) $\displaystyle f(x)=x^3,\ g(x)=\frac{1}{\sqrt[3]{x}}$

f) $\displaystyle f(x)=\frac{x}{1+x^2},\ g(x)=\frac{1}{x},$

g) $\displaystyle f(x)=\log x,\ g(x)=x^2+1.$

In [14]:

#a
forget()
x1,x2 = var('x1,x2')
assume(x1==-x2)

f(x) = abs(x)
g(x) = -3*x+2

ff(x) = f(f(x))
show(ff(x))
#x nalezy do R
#y nalezy do +R
fg(x) = f(g(x))
show(fg(x))
#x nalezy do R
#y nalezy do R
gf(x) = g(f(x))
show(gf(x))
#x nalezy do R
#y nalezy do R>-1
gg(x) = g(g(x))
show(gg(x))

$\displaystyle {\left| x \right|}$

$\displaystyle {\left| -3 \, x + 2 \right|}$

$\displaystyle -3 \, {\left| x \right|} + 2$

$\displaystyle 9 \, x - 4$

In [8]:

#b
forget()
x1,x2 = var('x1,x2')
assume(x1==-x2)

f(x) = sqrt(x)
g(x) = x^3 + 1

ff(x) = f(f(x))
show(ff(x))

fg(x) = f(g(x))
show(fg(x))

gf(x) = g(f(x))
show(gf(x))

gg(x) = g(g(x))
show(gg(x))

$\displaystyle x^{\frac{1}{4}}$

$\displaystyle \sqrt{x^{3} + 1}$

$\displaystyle x^{\frac{3}{2}} + 1$

$\displaystyle {\left(x^{3} + 1\right)}^{3} + 1$

In [13]:

#c
forget()
x1,x2 = var('x1,x2')
assume(x1==-x2)

f(x) = x^2
g(x) = sqrt(x)

ff(x) = f(f(x))
show(ff(x))

fg(x) = f(g(x))
show(fg(x))

gf(x) = g(f(x))
show(gf(x))

gg(x) = g(g(x))
show(gg(x))

$\displaystyle x^{4}$

$\displaystyle x$

$\displaystyle \sqrt{x^{2}}$

$\displaystyle x^{\frac{1}{4}}$

In [11]:

#d
forget()
x1,x2 = var('x1,x2')
assume(x1==-x2)

f(x) = 2**x
g(x) = cos(x)

ff(x) = f(f(x))
show(ff(x))

fg(x) = f(g(x))
show(fg(x))

gf(x) = g(f(x))
show(gf(x))

gg(x) = g(g(x))
show(gg(x))

$\displaystyle 2^{\left(2^{x}\right)}$

$\displaystyle 2^{\cos\left(x\right)}$

$\displaystyle \cos\left(2^{x}\right)$

$\displaystyle \cos\left(\cos\left(x\right)\right)$

In [10]:

#e ???? potegasx
forget()
x1,x2 = var('x1,x2')
assume(x1==-x2)

f(x) = x^3
g(x) = 1/(x**(1/3))

ff(x) = f(f(x))
show(ff(x))

fg(x) = f(g(x))
show(fg(x))

gf(x) = g(f(x))
show(gf(x))

gg(x) = g(g(x))
show(gg(x))

$\displaystyle x^{9}$

$\displaystyle \frac{1}{x}$

$\displaystyle \frac{1}{{\left(x^{3}\right)}^{\frac{1}{3}}}$

$\displaystyle x^{\frac{1}{9}}$

In [3]:

#f
forget()
x = var('x')

f(x) = x/(1+x^2)
g(x) = 1/x

ff(x) = f(f(x))
show(ff(x))

fg(x) = f(g(x))
show(fg(x))

gf(x) = g(f(x))
show(gf(x))

gg(x) = g(g(x))
show(gg(x))

$\displaystyle \frac{x}{{\left(x^{2} + 1\right)} {\left(\frac{x^{2}}{{\left(x^{2} + 1\right)}^{2}} + 1\right)}}$

$\displaystyle \frac{1}{x {\left(\frac{1}{x^{2}} + 1\right)}}$

$\displaystyle \frac{x^{2} + 1}{x}$

$\displaystyle x$

In [15]:

#g
forget()
x1,x2 = var('x1,x2')
assume(x1==-x2)

f(x) = log(x)
g(x) = x^2 + 1

ff(x) = f(f(x))
show(ff(x))
#x nalezy do +R (0,niesk) i log(x) > 0 wiec x > 1 (1,niesk)
fg(x) = f(g(x))
show(fg(x))
#x nalezy do R
gf(x) = g(f(x))
show(gf(x))
#x nalezy do R (0,niesk)
gg(x) = g(g(x))
show(gg(x))
#x nalezy do R

$\displaystyle \log\left(\log\left(x\right)\right)$

$\displaystyle \log\left(x^{2} + 1\right)$

$\displaystyle \log\left(x\right)^{2} + 1$

$\displaystyle {\left(x^{2} + 1\right)}^{2} + 1$

Zad. 6. Znaleźć funkcję odwrotną do podanej:

a) $\displaystyle f(x)=x^2-2x,\ x\in\langle 1;\infty)$

b) $\displaystyle g(x)=2-\sqrt[5]{x+1},\ x\in\mathbb{R}$

c) $\displaystyle h(x)=x^3|x|,\ x\in\mathbb{R},$

d) $\displaystyle p(x)=\left\{ \begin{array}{rrr} 3^x & \mbox{ dla } &x<0\\ 5^x & \mbox{ dla } &x\geqslant 0 \end{array}\right. ,\ x\in\mathbb{R}$

e) $\displaystyle f(x)=1-3^{-x}$

f) $\displaystyle g(x)=x^5+\sqrt{3},$

g) $\displaystyle h(x)=x^6\mathrm{sgn\,}x$

h) $\displaystyle q(x)=\left\{ \begin{array}{rrr} -x^2 & \mbox{ dla } &x<0\\ 2+x & \mbox{ dla } &x\geqslant 0 \end{array}\right. ,\ x\in\mathbb{R}$

i) $\displaystyle f(x)=\frac{x}{1+|x|}.$

In [19]:

#a 
forget()
x,y = var('x,y')
assume(x>1)

f(x) = x^2 - 2*x
show(solve(y==f(x),x))
#f do -1 (x) = sqrt(x+1)+1 dla x>=-1

$\displaystyle \left[x = \sqrt{y + 1} + 1\right]$

In [11]:

#b potega z wykladnikiem ? 
forget()
x,y = var('x,y')
assume(x>1)

f(x) = 2  - (x+1)**1/5
show(solve(y==f(x),x))
#f do -1 (x) = sqrt(x+1)+1 dla x>=-1

$\displaystyle \left[x = -5 \, y + 9\right]$

In [9]:

#c
forget()
x,y = var('x,y')

f(x) = (x^3 * abs(x))
show(solve(y==f(x),x))

$\displaystyle \left[x^{3} = \frac{y}{{\left| x \right|}}\right]$

In [6]:

#d
forget()
x = var('x')

assume(x<0)
p1(x) = 3^x

show(solve(y==p1(x),x))
forget()
assume(x>0)
p2(x) = 5^x
show(solve(y==p2(x),x))
#y=log trojkowy x, y<1
#analogicznie

$\displaystyle \left[x = \frac{\log\left(y\right)}{\log\left(3\right)}\right]$

$\displaystyle \left[x = \frac{\log\left(y\right)}{\log\left(5\right)}\right]$

In [12]:

#e
forget()
x,y = var('x,y')

f(x) = 1 - 3^-x
show(solve(y==f(x),x))

$\displaystyle \left[x = \frac{\log\left(-\frac{1}{y - 1}\right)}{\log\left(3\right)}\right]$

In [13]:

#f
forget()
x,y = var('x,y')

f(x) = x^5 + sqrt(3)
show(solve(y==f(x),x))

$\displaystyle \left[x = \frac{1}{4} \, {\left(y - \sqrt{3}\right)}^{\frac{1}{5}} {\left(\sqrt{5} + i \, \sqrt{2 \, \sqrt{5} + 10} - 1\right)}, x = -\frac{1}{4} \, {\left(y - \sqrt{3}\right)}^{\frac{1}{5}} {\left(\sqrt{5} - i \, \sqrt{-2 \, \sqrt{5} + 10} + 1\right)}, x = -\frac{1}{4} \, {\left(y - \sqrt{3}\right)}^{\frac{1}{5}} {\left(\sqrt{5} + i \, \sqrt{-2 \, \sqrt{5} + 10} + 1\right)}, x = \frac{1}{4} \, {\left(y - \sqrt{3}\right)}^{\frac{1}{5}} {\left(\sqrt{5} - i \, \sqrt{2 \, \sqrt{5} + 10} - 1\right)}, x = {\left(y - \sqrt{3}\right)}^{\frac{1}{5}}\right]$

In [14]:

#g
forget()
x,y = var('x,y')

f(x) = x^6 + sign(x)
show(solve(y==f(x),x))

$\displaystyle \left[x = \frac{1}{2} \, {\left(y - \mathrm{sgn}\left(x\right)\right)}^{\frac{1}{6}} {\left(i \, \sqrt{3} + 1\right)}, x = \frac{1}{2} \, {\left(y - \mathrm{sgn}\left(x\right)\right)}^{\frac{1}{6}} {\left(i \, \sqrt{3} - 1\right)}, x = -{\left(y - \mathrm{sgn}\left(x\right)\right)}^{\frac{1}{6}}, x = -\frac{1}{2} \, {\left(y - \mathrm{sgn}\left(x\right)\right)}^{\frac{1}{6}} {\left(i \, \sqrt{3} + 1\right)}, x = -\frac{1}{2} \, {\left(y - \mathrm{sgn}\left(x\right)\right)}^{\frac{1}{6}} {\left(i \, \sqrt{3} - 1\right)}, x = {\left(y - \mathrm{sgn}\left(x\right)\right)}^{\frac{1}{6}}\right]$

In [15]:

#h
forget()
assume(x<0)
p1(x) = (-x)^2

show(solve(y==p1(x),x))
forget()
assume(x>0)
p2(x) = 2 + x
show(solve(y==p2(x),x))

$\displaystyle \left[x = -\sqrt{y}\right]$

$\displaystyle \left[x = y - 2\right]$

In [16]:

#i
forget()
x,y = var('x,y')

f(x) = x/(1+abs(x))
show(solve(y==f(x),x))

$\displaystyle \left[x = y {\left| x \right|} + y\right]$

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Analiza matematyczna - PS

Informatyka, sem.I, studia niestacjonarne I stopnia, 2023/24

Lista nr 1: Funkcje jednej zmiennej. Własności funkcji

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Analiza matematyczna - PS

Informatyka, sem.I, studia niestacjonarne I stopnia, 2023/24

Lista nr 1: Funkcje jednej zmiennej. Własności funkcji

Real-time collaboration for Jupyter Notebooks, Linux Terminals, LaTeX, VS Code, R IDE, and more,
all in one place. Commercial Alternative to JupyterHub.