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

1 Answer
Douglas K.
Sep 7, 2017

Given: y = sqrt( x- 3x^2)

The argument of the square root must be greater than or equal to 0:

x - 3x^2 >= 0

Becuase this quadratic is the equation of a parabola that opens downward we know that, if we find the roots of the quadratic, the quadratic will be greater than 0 between the roots.

x - 3x^2 = 0

x(1 - 3x) = 0

x = 0 and x = 1/3

This gives us the domain 0 <= x <=1/3

We know that the minimum for the range occurs at either root:

0 <= y

We know that the maximum for the range will occur halfway between the roots, x = 1/6

y = sqrt( 1/6- 3(1/6)^2)

y = sqrt( 6/36- 3/36)

y = sqrt3/6

This makes the range become 0 <= y <= sqrt3/6

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