Question

Qual o domínio da função?
F(x)=√x²-4x-5



Pfv rápido

Answer 1

$f_{(x)}=\sqrt{x^2-4x-5}$

Para f(x) = 0

Deste modo ter-se-á:

$\sqrt{x^2-4x-5}=0$

$(\cancel{\sqrt{x^2-4x-5})^2}=0^2$

${\color{blue}{x^2-4x-5=0}}$

$Coeficientes:\left\{\begin{array}{cc}a=1\\b=-4\\c=-5\\\end{array}\right$

Bhaskara:

$\large\boxed{\boxed{{x_{1,2}=\frac{-b\pm\sqrt{b^2-4*a*c}}{2*a}}}}}}$

$x_{1,2}=\frac{4\pm\sqrt{(-4)^2-4*1*(-5)}}{2*1}$

$x_{1,2}=\frac{4\pm\sqrt{16+20}}{2}=\frac{4\pm\sqrt{36}}{2}$

${\color{blue}{x_{1}=\frac{4+6}{2}=\frac{10}{2}=5}}$

${\color{blue}{x_{2}=\frac{4-6}{2}=\frac{-2}{2}=-2}}$

$Sol:\{\:-2\:;\:5\}$

Domínio: |R\{-2 ; 5 }

Espero ter ajudado bastante:)