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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