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

Question

Genesis Allen · Accepted Answer

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

The function is

#y=sqrt(x-3x^2)#

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

Therefore,

#x-3x^2>=0#

#=>#, #3x^2-x<=0#

#=>#, #x(3x-1)<=0#

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

# x in [0, 1/3]#

The domain is # x in [0, 1/3]#

When #x=0#, #=>#, #y=0#

When #x=1/3#, #=>#, #y=0#

#y=sqrt(x-3x^2)#

#=>#, #y^2=x-3x^2#

#3x^2-x+y^2=0#

This is a quadratic equation in #x# and in order to have solutions, the discriminant #>=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]}

Eva Adamson · Answer

undefined To find the domain and range of $ \sqrt{x - 3x^2} $, we need to determine the values of $ x $ for which the expression under the square root is valid and the possible output values of the square root function.

For the domain:
1. The expression under the square root must be non-negative, so $ x - 3x^2 \geq 0 $.
2. Solve the inequality $ x - 3x^2 \geq 0 $ to find the valid values of $ x $.

For the range:
1. The square root function returns non-negative values, so the range will be all non-negative real numbers.

After solving the inequality for the domain, we will have the domain of the function. The range will be all non-negative real numbers (including zero).

Let's solve the inequality:
$ x - 3x^2 \geq 0 $
$ -3x^2 + x \geq 0 $
$ -x(3x - 1) \geq 0 $

This inequality holds when either both factors are non-positive or both factors are non-negative.

So, we have:
$ -x \geq 0 $ and $ 3x - 1 \geq 0 $

Solving each inequality separately:
1. For $ -x \geq 0 $, we have $ x \leq 0 $.
2. For $ 3x - 1 \geq 0 $, we have $ x \geq \frac{1}{3} $.

So, the domain of $ \sqrt{x - 3x^2} $ is $ x \leq 0 $ and $ x \geq \frac{1}{3} $.

The range is all non-negative real numbers, i.e., $ [0, \infty) $.