$a)f^{{}^{\prime }}\left(x\right)={\left(-x^3+3x^2+9x+2\right)}^{{}^{\prime }}={\left(-x^3\right)}^{{}^{\prime }}+{\left(3x^2\right)}^{{}^{\prime }}+{\left(9x\right)}^{{}^{\prime }}+{\left(2\right)}^{{}^{\prime }}=-3x^2+6x+9=-3\left(x^2-2x-3\right)=$  

$-3\left(x^2+x-3x-3\right)=-3\left(x\left(x+1\right)-3\left(x+1\right)\right)=-3\left(x+1\right)\left(x-3\right)$  

$f^{{}^{\prime }}\left(x\right)=0\text{gdy}-3\left(x+1\right)\left(x-3\right)=0$  

$x_1=-1\vee x_2=3$  

Wykresem funkcji jest parabola. Ramiona paraboli są skierowane ku dołowi a więc

$f^{{}^{\prime }}\left(x\right)<0\text{Dla}x\in \left(-\infty ,-1\right)\cup \left(3,\infty \right)$  

$f^{{}^{\prime }}\left(x\right)>0\text{Dla}x\in \left(-1,3\right)$  

A więc funkcja f ma ekstremum w punktach -1 , 3.

Minimum:

$f\left(-1\right)=-{\left(-1\right)}^3+3\cdot {\left(-1\right)}^2+9\cdot \left(-1\right)+2=1+3-9+2=-3$  

Maksimum:

$f\left(3\right)=-3^3+3\cdot 3^2+9\cdot 3+2=-27+27+27+2=29$  

 

$b)f^{{}^{\prime }}\left(x\right)={\left(x^4-8x^2+6\right)}^{{}^{\prime }}={\left(x^4\right)}^{{}^{\prime }}-{\left(8x^2\right)}^{{}^{\prime }}+{\left(6\right)}^{{}^{\prime }}=4x^3-16x=4x\left(x^2-4\right)=4x\left(x-2\right)\left(x+2\right)$  

$f^{{}^{\prime }}\left(x\right)=0\text{gdy}4x\left(x-2\right)\left(x+2\right)=0$  

$x_1=0\vee x_2=2\vee x_3=-2$  

 

	$\left(-\infty ,-2\right)$	$-2$	$\left(-2;0\right)$	$0$	$\left(0;2\right)$	$2$	$\left(2;\infty \right)$
$x+2$	$-$	$0$	$+$	$+$	$+$	$+$	$+$
$x$	$-$	$-$	$-$	$0$	$+$	$+$	$+$
$x-2$	$-$	$-$	$-$	$-$	$-$	$0$	$+$
znak	$-$	$0$	$+$	$0$	$-$	$0$	$+$

W każdym punkcie, gdzie wartość pochodnej wynosi 0, dochodzi do zmiany znaku pochodnej a więc funkcja f ma ekstremum w punktach -2, 0, 2.

Minimum:

$f\left(-2\right)={\left(-2\right)}^4-8\cdot {\left(-2\right)}^2+6=16-8\cdot 4+6=16-32+6=-10$  

$f\left(2\right)=2^4-8\cdot 2^2+6=16-8\cdot 4+6=16-32+6=-10$  

Maksimum:

$f\left(0\right)=0-8\cdot 0+6=6$  

 

$c)f^{{}^{\prime }}\left(x\right)={\left(\frac{1}{2}{x}^4+x^2-3\right)}^{{}^{\prime }}={\left(\frac{1}{2}{x}^4\right)}^{{}^{\prime }}+{\left(x^2\right)}^{{}^{\prime }}-{\left(3\right)}^{{}^{\prime }}=\frac{1}{2}\cdot 4x^3+2x=2x^3+2x=2x\left(x^2+1\right)$  

$f^{{}^{\prime }}\left(x\right)=0\text{gdy}2x\left(x^2+1\right)=0$  

$2x\left(x^2+1\right)=0$  

$2x=0$  

$x_1=0$  

Na lewo od punktu 0 wartość pochodnej jest ujemna, na prawo dodatnia a więc w punkcie 0 funkcja f ma ekstremum, jest to minimum.

$f\left(0\right)=\frac{1}{2}\cdot 0+0-3=-3$  

 

$d)f^{{}^{\prime }}\left(x\right)={\left(x^5+x^2-2\right)}^{{}^{\prime }}={\left(x^5\right)}^{{}^{\prime }}+{\left(x^2\right)}^{{}^{\prime }}-{\left(2\right)}^{{}^{\prime }}=5x^4+2x=x\left(5x^3+2\right)$  

$f^{{}^{\prime }}\left(x\right)=0\text{gdy}x\left(5x^3+2\right)=0$  

$x\left(5x^3+2\right)=0$  

$x_1=0\vee 5x^3+2=0$  

$x_1=0\vee x^3=-\frac{2}{5}$  

$x_1=0\vee x_2=-\sqrt[3]{\frac{2}{5}}$

 

	$\left(-\infty ,-\sqrt[3]{\frac{2}{5}}\right)$	$-\sqrt[3]{\frac{2}{5}}$	$\left(-\sqrt[3]{\frac{2}{5}};0\right)$	$0$	$\left(0;\infty \right)$
$\left(5x^3+2\right)$	$-$	$0$	$+$	$+$	$+$
$x$	$-$	$-$	$-$	$0$	$+$
znak	$+$	$0$	$-$	$0$	$+$

A więc funkcja ma ekstremum w punktach:

$-\sqrt[3]{\frac{2}{5}},0$  

Maksimum:

$f\left(-\sqrt[3]{\frac{2}{5}}\right)={\left(-\sqrt[3]{\frac{2}{5}}\right)}^5+{\left(-\sqrt[3]{\frac{2}{5}}\right)}^2-2=-\sqrt[3]{{\left(\frac{2}{5}\right)}^3{\left(\frac{2}{5}\right)}^2}+\sqrt[3]{{\left(\frac{2}{5}\right)}^2}-2=-\frac{2}{5}\cdot \sqrt[3]{\frac{4}{25}}+\sqrt[3]{\frac{4}{25}}-2=\frac{3}{5}\sqrt[3]{\frac{4}{25}}-2$  

Minimum:

$f\left(0\right)=0+0-2=-2$  

 

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

$f^{{}^{\prime }}\left(x\right)=0\text{gdy}x-\frac{1}{x^2}=0$  

$x-\frac{1}{x^2}=0\mid \cdot x^2$  

$x^3-1=0$  

$x^3=1$  

$x_1=1$  

Na lewo od punktu 1 wartość pochodnej jest ujemna, natomiast na prawo jest dodatnia a więc funkcja ma ekstremum w punkcie 1. Jest to minimum.

$f\left(1\right)=\frac{1}{2}\cdot 1+\frac{1}{1}=\frac{1}{2}+1=\frac{3}{2}$  

 

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

$f^{{}^{\prime }}\left(x\right)=0\text{gdy}3x^2-\frac{3}{x^2}=0$  

$3x^2-\frac{3}{x^2}=0\mid \cdot x^2$  

$3x^4-3=0$  

$x^4-1=0$  

$\left(x^2-1\right)\left(x^2+1\right)=0$  

$\left(x-1\right)\left(x+1\right)\left(x^2+1\right)=0$  

$\left(x-1\right)\left(x+1\right)=0$  

$x_1=1\vee x_2=-1$  

 

	$\left(-\infty ,-1\right)$	$-1$	$\left(-1;1\right)$	$1$	$\left(1;\infty \right)$
$\left(x+1\right)$	$-$	$0$	$+$	$+$	$+$
$x-1$	$-$	$-$	$-$	$0$	$+$
znak	$+$	$0$	$-$	$0$	$+$

A więc funkcja ma ekstremum w punktach -1, 1.

Maksimum:

$f\left(-1\right)={\left(-1\right)}^3+\frac{3}{-1}=-1-3=-4$  

Minimum:

$f\left(1\right)=1^3+\frac{3}{1}=1+3=4$