How do you find domain and range for #f(x) =sqrt( x- (3x^2))#?

Question

Kennedy Adams · Accepted Answer

Refer to explanation

That is available for the domain. #x-3x^2>=0=>x*(1-3x)>=0# which is true for #0<=x<=1/3# Hence the domain is #[0,1/3]# For the range we set #y=f(x)>=0# so #y=sqrt(x-3x^2)=>y^2=x-3x^2=>3x^2-x+y^2=0# The discriminant for the last must be greater than or equal to zero because it is a trinomial with respect to x. #(-1)^2-4*3*(y^2)>=0=>1/12>=y^2=>1/(2sqrt3)>=y# Hence the range is #R(f)=[0,1/(2sqrt3)]#

Kennedy Adams · Answer

undefined To find the domain of $ f(x) = \sqrt{x - 3x^2} $, we need to identify the values of $ x $ for which the function is defined. Since the square root function is only defined for non-negative values, the expression inside the square root, $ x - 3x^2 $, must be greater than or equal to zero.

$ x - 3x^2 \geq 0 $

To find the values of $ x $ that satisfy this inequality, we solve for $ x $:

$ x(1 - 3x) \geq 0 $

The critical points are $ x = 0 $ and $ x = \frac{1}{3} $. Testing intervals around these critical points, we find that the domain of the function is $ x \in (-\infty, 0] \cup [\frac{1}{3}, \infty) $.

To find the range, we need to consider the possible output values of the function. Since the square root function outputs non-negative values, the range of $ f(x) = \sqrt{x - 3x^2} $ is $ y \geq 0 $.