Question

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

Answer 1

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`