what is the domain of the function f(x)=1/root[2x-3]
the domain is all reals except where the denominator
a) is zero
b) is undefined
√(2x-3) = 0 when x = 3/2
√(2x-3) is undefined when 2x-3<0, or x<3/2
So, the domain is all x > 3/2
To find the domain of a function, we need to determine the set of all possible values for which the function is defined.
In the given function, f(x) = 1/√(2x-3).
The square root function is defined for non-negative real numbers. This means that the value inside the square root (√) must be greater than or equal to zero.
So, 2x - 3 ≥ 0
Adding 3 to both sides, we get:
2x ≥ 3
Dividing both sides by 2, we get:
x ≥ 3/2
Therefore, the domain of the function f(x) = 1/√(2x-3) is all the real numbers greater than or equal to 3/2. In interval notation, the domain would be [3/2, +∞).
To find the domain of the function f(x) = 1/√(2x-3), we need to determine the values of x for which the function is defined.
For the given function, the expression inside the square root (√) must be greater than or equal to zero. This is because the square root of a negative number is undefined in the real number system.
So, we solve the inequality: 2x - 3 ≥ 0
Adding 3 to both sides of the inequality, we get:
2x ≥ 3
Dividing both sides by 2, we have:
x ≥ 3/2
Therefore, the domain of the function f(x) = 1/√(2x-3) is all real numbers such that x is greater than or equal to 3/2. In interval notation, the domain is [3/2, +∞).