Halle el dominio de la función f(x)=√(2x^2-5x+3)
CarlosMath
![\displaystyle 2x^2-5x+3\geq0\\ \\ (2x-3)(x-1)\geq0\\ \\ x\in(-\infty,1]\cup \left[\frac{3}{2},+\infty\right)\\ \\ \\ \text{Entonces Dom}f=(-\infty,1]\cup \left[\frac{3}{2},+\infty\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%0A2x%5E2-5x%2B3%5Cgeq0%5C%5C+%5C%5C%0A%282x-3%29%28x-1%29%5Cgeq0%5C%5C+%5C%5C%0Ax%5Cin%28-%5Cinfty%2C1%5D%5Ccup+%5Cleft%5B%5Cfrac%7B3%7D%7B2%7D%2C%2B%5Cinfty%5Cright%29%5C%5C+%5C%5C+%5C%5C%0A%5Ctext%7BEntonces+Dom%7Df%3D%28-%5Cinfty%2C1%5D%5Ccup+%5Cleft%5B%5Cfrac%7B3%7D%7B2%7D%2C%2B%5Cinfty%5Cright%29 "\displaystyle 2x^2-5x+3\geq0\\ \\ (2x-3)(x-1)\geq0\\ \\ x\in(-\infty,1]\cup \left[\frac{3}{2},+\infty\right)\\ \\ \\ \text{Entonces Dom}f=(-\infty,1]\cup \left[\frac{3}{2},+\infty\right)")
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