Patricia Bean

2022-07-17

I know that fraction denominator needs to be > 0 , so if denominator is quadractic equation i know how to solve , But since denominator is > 0 , I dont know how to solve , can anyone help me ?
$f\left(x\right)=\sqrt{\mathrm{log}\left(\frac{3x-{x}^{2}}{2}\right)}$

umgangistbf

Hints:
You also need to have
$\mathrm{log}\left(\frac{3x-{x}^{2}}{2}\right)\ge 0$
for the square root of that function to be defined, and this implies
$\frac{3x-{x}^{2}}{2}\ge 1\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}3x-{x}^{2}\ge 2$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}3x-{x}^{2}-2\ge 0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{x}^{2}-3x+2=\left(x-1\right)\left(x-2\right)\le 0.$
The quadratic will be less than or equal to zero when exactly one and only one factor $\le 0$:
So the domain is, in fact,

Mauricio Mathis

it must be
$\frac{3x-{x}^{2}}{2}\ge 1$
this is equivalent to $0\ge {x}^{2}-3x+2$ and this is equivalent to
$0\ge \left(x-1\right)\left(x-2\right)$

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