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)]#