State the domain and range of the function
f(x) = 2sin(sqrt(3x-1)) + 3
Why is this so difficult?
no idea. Just take it step by step.
domain of √x is x>=0
so, the domain of √(3x-1) is 3x-1 >= 0, or x >= 1/3
the domain of sin(x) is all reals, so
the domain of sin(√(3x-1)) is the domain of √(3x-1), or x >= 1/3
of course, the range is just the same as the range of 2sin(x)+3: [1,5]
http://www.wolframalpha.com/input/?i=2sin(sqrt(3x-1))+%2B+3
Determining the domain and range of a function can sometimes be challenging due to the complexity of the function's equation or restrictions on the possible values of the independent variable. However, with some analysis using the given function, we can identify the domain and range.
To find the domain of a function, we need to consider any restrictions on the possible values for the independent variable (x). In this case, there are no specific restrictions mentioned in the equation. However, we need to consider any potential limitations on x that could make the function undefined.
For the given function f(x) = 2sin(sqrt(3x-1)) + 3, the square root function is defined only for non-negative values. Therefore, the expression inside the square root (3x - 1) must be greater than or equal to zero:
3x - 1 ≥ 0
Solving the inequality, we find:
3x ≥ 1
x ≥ 1/3
So the domain of the function is all real numbers greater than or equal to 1/3, or in interval notation, [1/3, ∞).
Now, let's determine the range of the function. The sine function outputs values between -1 and 1. The entire function f(x) = 2sin(sqrt(3x-1)) + 3 is the sum of these sine values and three, so the minimum value of the function is 3 - 1 = 2, and the maximum value is 3 + 1 = 4.
Therefore, the range of the function is [2, 4].
In conclusion, the domain of the function f(x) = 2sin(sqrt(3x-1)) + 3 is [1/3, ∞) and the range is [2, 4].