How do you find the domain and range of #sqrt ( x- (3x^2))#?

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Answer 1 · 2018-08-11T16:36:02

The domain is # x in [0, 1/3]#. The range is #y in [0, 0.289]#

The function is

#y=sqrt(x-3x^2)#

What's under the square root sign is #>=0#

Therefore,

#x-3x^2>=0#

#=>#, #3x^2-x<=0#

#=>#, #x(3x-1)<=0#

The solution to this inequality (obtained with a sign chart) is

# x in [0, 1/3]#

The domain is # x in [0, 1/3]#

When #x=0#, #=>#, #y=0#

When #x=1/3#, #=>#, #y=0#

#y=sqrt(x-3x^2)#

#=>#, #y^2=x-3x^2#

#3x^2-x+y^2=0#

This is a quadratic equation in #x# and in order to have solutions, the discriminant #>=0#

#Delta=(-1)^2-4(3)(y^2)>=0#

#1-12y^2>=0#

#y^2<=1/12#

#y<=+-sqrt(1/12)#

We keep the positive solution

#y<=1/sqrt(12)<=0.289#

The range is #y in [0, 0.289]#

graph{sqrt(x-3x^2) [-0.192, 0.5473, -0.044, 0.3256]}