a)

$f\left(x\right)=\frac{1}{5}{x}^5-\frac{10}{3}{x}^3+9x-10$  

$f{}^{\prime }\left(x\right)=\frac{1}{5}\cdot 5x^4-\frac{10}{3}\cdot 3x^2+9$  

$f{}^{\prime }\left(x\right)=x^4-10x^2+9$  

Sprawdźmy kiedy f'(x)=0.

$x^4-10x^2+9=0$  

Podstawmy  $x^2=t$  

$t^2-10t+9=0$  

$\Delta ={\left(-10\right)}^2-4\cdot 1\cdot 9=100-36=64$  

$t_1=\frac{10-8}{2}=\frac{2}{2}=1$  

$t_2=\frac{10+8}{2}=\frac{18}{2}=9$  

$x^2=1\vee x^2=9$  

$x\in \left\{-1,1,-3,3\right\}$  

Badamy kiedy f'(x)>0. 

$\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(x+3\right)>0$  

$x\in \left(-\infty ,-3\right)\cup \left(-1,1\right)\cup \left(3,+\infty \right)$  

 

Budujemy tabelkę.

$x$	$\left(-\infty ,-3\right)$	$-3$	$\left(-3,-1\right)$	$-1$	$\left(-1,1\right)$	$1$	$\left(1,3\right)$	$3$	$\left(3,+\infty \right)$
$f\left(x\right)$	rosnąca	maksimum lokalne	malejąca	minimum lokalne	rosnąca	maksimum lokalne	malejąca	minimum lokalne	rosnąca

$f_{\max }\left(-3\right)=\frac{1}{5}\cdot {\left(-3\right)}^5-\frac{10}{3}\cdot {\left(-3\right)}^3+9\cdot \left(-3\right)-10=-\frac{243}{5}+90-27-10=-48\frac{3}{5}+53=4\frac{2}{5}$  

$f_{\min }\left(-1\right)=\frac{1}{5}\cdot {\left(-1\right)}^5-\frac{10}{3}\cdot {\left(-1\right)}^3+9\cdot \left(-1\right)-10=-\frac{1}{5}+\frac{10}{3}-9-10=-\frac{3}{15}+\frac{50}{15}-19=\frac{47}{15}-19=3\frac{2}{15}-19=-15\frac{13}{15}$  

$f_{\max }\left(1\right)=\frac{1}{5}-\frac{10}{3}+9-10=\frac{3}{15}-\frac{50}{15}-1=-\frac{47}{15}-1=-3\frac{2}{15}-1=-4\frac{2}{15}$  

$f_{\min }\left(3\right)=\frac{1}{5}\cdot 3^5-\frac{10}{3}\cdot 3^3+9\cdot 3-10=\frac{243}{5}-90+27-10=48\frac{3}{5}-73=-24\frac{2}{5}$  

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b)

$f\left(x\right)=\frac{1}{4}{x}^4-2x^3+4\frac{1}{2}{x}^2+7$  

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

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

Sprawdźmy kiedy f'(x)=0.

$x^3-6x^2+9x=0$  

$x\left(x^2-6x+9\right)=0$  

$x{\left(x-3\right)}^2=0$  

$x=0\vee x=3$  

Badamy kiedy f'(x)>0.

$x{\left(x-3\right)}^2>0$  

$x>0$  

 

$f\left(3\right)=\frac{1}{4}\cdot 3^4-2\cdot 3^3+\frac{9}{2}\cdot 3^2+7$  

$f\left(3\right)=\frac{1}{4}\cdot 81-54+\frac{81}{2}+7$  

$f\left(3\right)=\frac{81}{4}-54+\frac{162}{4}+7$  

$f\left(3\right)=\frac{55}{4}$  

Budujemy tabelkę.

$x$	$\left(-\infty ,0\right)$	$0$	$\left(0,3\right)$	$3$	$\left(3,+\infty \right)$
$f\left(x\right)$	malejąca	minimum lokalne	rosnąca	$\frac{55}{4}$	rosnąca

$f_{\min }\left(0\right)=\frac{1}{4}\cdot 0^4-2\cdot 0^3+4\frac{1}{2}\cdot 0^2+7=7$  

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c)

$f\left(x\right)=\frac{1}{5}{x}^5-x^4-4x^3+84$  

$f{}^{\prime }\left(x\right)=\frac{1}{5}\cdot 5x^4-4x^3-4\cdot 3x^2$  

$f{}^{\prime }\left(x\right)=x^4-4x^3-12x^2$  

Sprawdźmy kiedy f'(x)=0. 

$x^4-4x^3-12x^2=0$  

$x^2\left(x^2-4x-12\right)=0$  

$\Delta ={\left(-4\right)}^2-4\cdot 1\cdot \left(-12\right)=16+48=64$  

$x_1=\frac{4-8}{2}=\frac{-4}{2}=-2$  

$x_2=\frac{4+8}{2}=\frac{12}{2}=6$  

$x=0\vee x=-2\vee x=6$  

Badamy kiedy f'(x)>0.

$x^2\left(x+2\right)\left(x-6\right)>0$  

$\left(x+2\right)\left(x-6\right)>0$  

$x\in \left(-\infty ,-2\right)\cup \left(6,+\infty \right)$  

 

$f\left(0\right)=84$  

Budujemy tabelkę.

$x$	$\left(-\infty ,-2\right)$	$-2$	$\left(-2,0\right)$	$0$	$\left(0,6\right)$	$6$	$\left(6,+\infty \right)$
$f\left(x\right)$	rosnąca	maksimum lokalne	malejąca	$84$	malejąca	minimum lokalne	rosnąca

$f_{\max }\left(-2\right)=\frac{1}{5}\cdot {\left(-2\right)}^5-{\left(-2\right)}^4-4\cdot {\left(-2\right)}^3+84={\left(-2\right)}^3\cdot \left(\frac{1}{5}\cdot {\left(-2\right)}^2+2-4\right)+84=\left(-8\right)\cdot \left(\frac{4}{5}-2\right)+84=\left(-8\right)\cdot \left(-1,2\right)+84=9,6+84=93,6$  

$f_{\min }\left(6\right)=\frac{1}{5}\cdot 6^5-6^4-4\cdot 6^3+84=6^3\left(\frac{1}{5}\cdot 6^2-6-4\right)+84=216\cdot \left(\frac{36}{5}-10\right)+84=216\cdot \left(7,2-10\right)+84=216\cdot \left(-2,8\right)+84=-604,8+84=-520,8$  

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d)

$f\left(x\right)=\frac{1}{5}{x}^5+3x^3+18x-123$  

$f{}^{\prime }\left(x\right)=\frac{1}{5}\cdot 5x^4+3\cdot 3x^2+18$  

$f{}^{\prime }\left(x\right)=x^4+9x^2+18$  

Sprawdźmy kiedy f'(x)=0.

$x^4+9x^2+18=0$  

Podstawmy  $x^2=t$  

$t^2+9t+18=0$  

$\Delta =9^2-4\cdot 1\cdot 18=81-72=9$  

$t_1=\frac{-9-3}{2}=\frac{-12}{2}=-6$  

$t_2=\frac{-9+3}{2}=\frac{-6}{2}=-3$  

$x^2=-6\vee x^2=-3$  

$x\in \varnothing$  

Brak ekstremów lokalnych.