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 ?

$f\left(x\right)=\sqrt{\mathrm{log}\left(\frac{3x-{x}^{2}}{2}\right)}$

$f\left(x\right)=\sqrt{\mathrm{log}\left(\frac{3x-{x}^{2}}{2}\right)}$

umgangistbf

Beginner2022-07-18Added 12 answers

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

$\frac{3x-{x}^{2}}{2}\ge 1\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}3x-{x}^{2}\ge 2$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}3x-{x}^{2}-2\ge 0\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{x}^{2}-3x+2=(x-1)(x-2)\le 0.$

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,

$x\in [1,2],\phantom{\rule{thickmathspace}{0ex}}\text{alternatively}\phantom{\rule{thinmathspace}{0ex}}1\le x\le 2$

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

$\frac{3x-{x}^{2}}{2}\ge 1\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}3x-{x}^{2}\ge 2$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}3x-{x}^{2}-2\ge 0\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{x}^{2}-3x+2=(x-1)(x-2)\le 0.$

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,

$x\in [1,2],\phantom{\rule{thickmathspace}{0ex}}\text{alternatively}\phantom{\rule{thinmathspace}{0ex}}1\le x\le 2$

Mauricio Mathis

Beginner2022-07-19Added 2 answers

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

$0\ge (x-1)(x-2)$

$\frac{3x-{x}^{2}}{2}\ge 1$

this is equivalent to $0\ge {x}^{2}-3x+2$ and this is equivalent to

$0\ge (x-1)(x-2)$

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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