Question

Determine o domínio da função:

Answer 1

Resposta:

x≠4 e x≠1

Explicação passo a passo:

5x²-12x+7≠0

$\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~5x^{2}-12x+7=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~a=5{;}~b=-12~e~c=7\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-12)^{2}-4(5)(7)=144-(140)=4\\\\C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-12)-\sqrt{4}}{2(5)}=\frac{12-2}{10}=\frac{10}{10}=1$

$\displaystyle x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-12)+\sqrt{4}}{2(5)}=\frac{12+2}{10}=\frac{14\div2}{10\div2}=\frac{7}{5}$