Please help me. Please show work.
What are the domain and range of the function y=2 square root 3x + 4-5?
*A.) X >/= -4/3; Y >/=-5
B.) X >/= 4/3; Y >/= 5
C.) X </= -4/3; Y </= -5
D.) X </= 4/3; Y </= 5
Thank You so much.
for √ to be defined, we need
3x+4 >= 0
3x >= -4
x >= -4/3
At x = -4/3, y = -5, so for any x >= -4/3, y >= -5
Looks like (A) all right
THANK YOU SO SO SO MUCH!!!!!!!
@Steve is amazing
To find the domain and range of the function y = 2 sqrt(3x) + 4 - 5, we need to consider the restrictions on the values of x and the resulting values of y.
Domain:
In this function, there is a square root of 3x. For the square root to be defined, the argument (3x) must be greater than or equal to zero. So we set up the inequality:
3x ≥ 0
Dividing both sides by 3, we have:
x ≥ 0
Therefore, the domain of the function is x ≥ 0.
Range:
To determine the range, we need to analyze the possible values of y. Since the function is a square root function, the values of y will be greater than or equal to 0.
The given function also has the constant terms (+4 - 5), which will not affect the range since they are added/subtracted after computing the square root. Hence, we only need to consider the square root part, which is always non-negative.
So, the range of the function is y ≥ 0.
Now, let's analyze the answer choices:
A.) x ≥ -4/3; y ≥ -5
B.) x ≥ 4/3; y ≥ 5
C.) x ≤ -4/3; y ≤ -5
D.) x ≤ 4/3; y ≤ 5
Comparing the answer choices with our findings:
- Choice A.) is incorrect because the domain should include x = 0.
- Choice B.) is incorrect because the domain should include x = 0.
- Choice C.) is incorrect because the range should include y = 0.
- Choice D.) matches our findings, which is x ≥ 0 and y ≥ 0.
Therefore, the correct answer is D.) x ≤ 4/3; y ≤ 5.
Please note that the inequality signs in the answer choices were reversed; they should have been "greater than or equal to," but the correct conclusion can still be derived by interpreting them correctly.