How do you find the domain and range of sqrt ( x- (3x^2))√x−(3x2)?

1 Answer
Narad T.
Aug 11, 2018

The domain is x in [0, 1/3]x∈[0,13]. The range is y in [0, 0.289]y∈[0,0.289]

Explanation:

The function is

y=sqrt(x-3x^2)y=√x−3x2

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

Therefore,

x-3x^2>=0x−3x2≥0

=>⇒, 3x^2-x<=03x2−x≤0

=>⇒, x(3x-1)<=0x(3x−1)≤0

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

x in [0, 1/3]x∈[0,13]

The domain is x in [0, 1/3]x∈[0,13]

When x=0x=0, =>⇒, y=0y=0

When x=1/3x=13, =>⇒, y=0y=0

y=sqrt(x-3x^2)y=√x−3x2

=>⇒, y^2=x-3x^2y2=x−3x2

3x^2-x+y^2=03x2−x+y2=0

This is a quadratic equation in xx and in order to have solutions, the discriminant >=0≥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]}

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