What is the domain of f(x)=sqrt(2x^2-5x) ?

The domain of f(x)=sqrt(2x^2-5x) is x = 5/2

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domain of f(x)=sqrt(2x^2-5x)

domain

f (x) = 2 x^{2} - 5 x

x \leq 0 or x \geq \frac{5}{2}

Interval Notation

(- \infty, 0] \cup [\frac{5}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq 0 or x \geq \frac{5}{2}

The function domain

x \leq 0 or x \geq \frac{5}{2}

f (x) = 4 - 2 x

f (x) = 4 x^{2} + 5 x - 3

f (x) = 2 ln (x - 1) + 5

f (x) = - 2 x + 7

y = 6 cos (2 π x)

What is the domain of f(x)=sqrt(2x^2-5x) ?
The domain of f(x)=sqrt(2x^2-5x) is x<= 0\lor x>= 5/2