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In order for f(x) to be a real number, the expression (4x^2 - 5) must be 0 or greater. If that number equals less than zero, its square root will be imaginary.
So x can only be values such that (4x^2 - 5) >= 0.
4x^2 - 5 >= 0
4x^2 >= 5
x^2 >= 5/4
x >= sqrt(5/4), or sqrt(5) / 2
That is the domain of the function.
The range will be real numbers.
Domain is:
x^2 >= 1.25
x >= sqrt(1.25) or x <= -sqrt(1.25)
Approximately:
x >= 1.1180339887498948482045868343656 or x <= -1.1180339887498948482045868343656
The range is: f(x) >= 0, since sqrt() is assumed to yield a non-negative number unless otherwise specified.
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Verified answer
In order for f(x) to be a real number, the expression (4x^2 - 5) must be 0 or greater. If that number equals less than zero, its square root will be imaginary.
So x can only be values such that (4x^2 - 5) >= 0.
4x^2 - 5 >= 0
4x^2 >= 5
x^2 >= 5/4
x >= sqrt(5/4), or sqrt(5) / 2
That is the domain of the function.
The range will be real numbers.
Domain is:
4x^2 - 5 >= 0
4x^2 >= 5
x^2 >= 5/4
x^2 >= 1.25
x >= sqrt(1.25) or x <= -sqrt(1.25)
Approximately:
x >= 1.1180339887498948482045868343656 or x <= -1.1180339887498948482045868343656
The range is: f(x) >= 0, since sqrt() is assumed to yield a non-negative number unless otherwise specified.