How do you solve $((x-1)(3-x))/(x-2)^2>=0$ ?

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Answer 1 · 2016-11-30T09:15:49

$f(x)≥0$ for $1≤x≤3$ with the exclusion of $x=2$ where it is undefined.

The denominator is always positive, so the sign of $f(x)$ is determined by the numerator.

The numerator is positive when both factors are positive or both are negative.

$(x-1) >= 0$ for $x>=1$
$(3-x) >=0$ for $x<=3$

So $f(x) >=0$ for $1<=x<=3$ with the exclusion of $x=2$ where it is undefined.

Answer 2 · 2016-11-30T10:22:29

The answer is $x in [ 1,2[ uu ]2, 3 ]$

Let $f(x)=((x-1)(3-x))/(x-2)^2$

The domain of $f(x)$ is $D_f(x)=RR-{2}$

Also the denominator is $>0$ , $AAx in D_f(x)$

So, we make the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $1$ $color(white)(aaaa)$ $3$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x-1$ $color(white)(aaaa)$ $-$ $color(white)(aaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $3-x$ $color(white)(aaaa)$ $+$ $color(white)(aaa)$ $+$ $color(white)(aaaa)$ $-$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaa)$ $-$ $color(white)(aaa)$ $+$ $color(white)(aaaa)$ $-$

So,

$f(x)>=0$ when $x in [ 1,2[ uu ]2, 3 ]$

graph{((x-1)(3-x))/(x-2)^2 [-12.66, 12.65, -6.33, 6.33]}

How do you solve ((x-1)(3-x))/(x-2)^2>=0((x-1)(3-x))/(x-2)^2>=0?