Dada a função f(x)=2x²-1, encontre caso exista suas raízes.
Soluções para a tarefa
Respondido por
1
veja...
![f(x)=2x^2-1 \\ \\ 2x^2-1=0 \\ \\ 2x^2=1 \\ \\ x^2= \frac{1}{2} \\ \\ x=\pm \sqrt{ \frac{1}{2} } \\ \\ x=\pm \frac{1}{ \sqrt{2} } \\ \\ racionalizando \\ \\ x=\pm \frac{1. \sqrt{2} }{ \sqrt{2}. \sqrt{2} } =\pm \frac{ \sqrt{2} }{ \sqrt{4} } =\pm \frac{ \sqrt{2} }{2} \\ \\ S=\{- \frac{ \sqrt{2} }{2} ,+ \frac{ \sqrt{2} }{2} \} f(x)=2x^2-1 \\ \\ 2x^2-1=0 \\ \\ 2x^2=1 \\ \\ x^2= \frac{1}{2} \\ \\ x=\pm \sqrt{ \frac{1}{2} } \\ \\ x=\pm \frac{1}{ \sqrt{2} } \\ \\ racionalizando \\ \\ x=\pm \frac{1. \sqrt{2} }{ \sqrt{2}. \sqrt{2} } =\pm \frac{ \sqrt{2} }{ \sqrt{4} } =\pm \frac{ \sqrt{2} }{2} \\ \\ S=\{- \frac{ \sqrt{2} }{2} ,+ \frac{ \sqrt{2} }{2} \}](https://tex.z-dn.net/?f=f%28x%29%3D2x%5E2-1+%5C%5C++%5C%5C+2x%5E2-1%3D0+%5C%5C++%5C%5C+2x%5E2%3D1+%5C%5C++%5C%5C+x%5E2%3D+%5Cfrac%7B1%7D%7B2%7D++%5C%5C++%5C%5C+x%3D%5Cpm+%5Csqrt%7B+%5Cfrac%7B1%7D%7B2%7D+%7D++%5C%5C++%5C%5C+x%3D%5Cpm+%5Cfrac%7B1%7D%7B+%5Csqrt%7B2%7D+%7D++%5C%5C++%5C%5C+racionalizando+%5C%5C++%5C%5C+x%3D%5Cpm+%5Cfrac%7B1.+%5Csqrt%7B2%7D+%7D%7B+%5Csqrt%7B2%7D.+%5Csqrt%7B2%7D++%7D+%3D%5Cpm+%5Cfrac%7B+%5Csqrt%7B2%7D+%7D%7B+%5Csqrt%7B4%7D+%7D+%3D%5Cpm+%5Cfrac%7B+%5Csqrt%7B2%7D+%7D%7B2%7D++%5C%5C++%5C%5C+S%3D%5C%7B-+%5Cfrac%7B+%5Csqrt%7B2%7D+%7D%7B2%7D+%2C%2B+%5Cfrac%7B+%5Csqrt%7B2%7D+%7D%7B2%7D+%5C%7D)
VitoriaEstevo16:
Obrigado
Perguntas interessantes