How do you find domain and range for $f(x) =sqrt( x- (3x^2))$ ?

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Answer 1 · 2015-09-25T06:22:17

Refer to explanation

We have that for the domain is

$x-3x^2>=0=>x*(1-3x)>=0$ which is true for $0<=x<=1/3$

Hence the domain is $[0,1/3]$

For the range we set $y=f(x)>=0$ so

$y=sqrt(x-3x^2)=>y^2=x-3x^2=>3x^2-x+y^2=0$

The last is a trinomial with respect to x so the discriminant must be greater or equal to zero hence

$(-1)^2-4*3*(y^2)>=0=>1/12>=y^2=>1/(2sqrt3)>=y$

Hence the range is $R(f)=[0,1/(2sqrt3)]$

How do you find domain and range for f(x) =sqrt( x- (3x^2))f(x) =sqrt( x- (3x^2))?