Não há uma solução fácil de ser calculada. Pela aproximação de Newton, obtemos (análise Numérica):
n ##### xn #################### f(xn) #################### f'(xn) #################### x(n+1)
1 ##### 5,000000000 ########## 2,236067977 ########## 10,2236068 ########## 4,781283844
2 ##### 4,781283844 ########## 0,047289898 ########## 9,791231659 ########## 4,776454023
3 ##### 4,776454023 ########## 2,30481E-05 ########## 9,781687597 ########## 4,776451667
4 ##### 4,776451667 ########## 5,48539E-12 ########## 9,781682941 ########## 4,776451667
5 ##### 4,776451667 ########## 0000000000 ########## 9,781682941 ########## 4,776451667
x = 4,776451667
Utilizado:
Eu faço a diferença. E você?
Do Poema: Quanto os professores "fazem"?
De Taylor Mali
![{x}^{2} + \sqrt[]{x} - 25 = 0](/latexrender/pictures/c9f3b898c1be36d626990445eb50831c.png "{x}^{2} + \sqrt[]{x} - 25 = 0")
por Claudio Matos » Ter Ago 18, 2015 15:43 
![{(0,05)}^{-\frac{1}{2}}=\frac{10}{\sqrt[5]}](/latexrender/pictures/19807748a214d3361336324f3e43ea9a.png "{(0,05)}^{-\frac{1}{2}}=\frac{10}{\sqrt[5]}")
![{(0,05)}^{-\frac{1}{2}}=\frac{10}{\sqrt[2]{5}}](/latexrender/pictures/3d7908e5b4e397bf635b6546063d9130.png "{(0,05)}^{-\frac{1}{2}}=\frac{10}{\sqrt[2]{5}}")
, ou seja, 1 dividido por 20 é igual a 0.05 . Sendo assim, a função final é igual a vinte elevado à meio. ![{0,05}^{-\frac{1}{2}} = {\frac{1}{20}}^{-\frac{1}{2}} = {\frac{20}{1}}^{\frac{1}{2}} = \sqrt[2]{20}](/latexrender/pictures/c0100c6f4d8bdbb7d54165e6be7aff04.png "{0,05}^{-\frac{1}{2}} = {\frac{1}{20}}^{-\frac{1}{2}} = {\frac{20}{1}}^{\frac{1}{2}} = \sqrt[2]{20}")
da seguinte forma:
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da seguinte forma:
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