What is the domain of f(x)=sqrt(2x^2+x-3) ?

The domain of f(x)=sqrt(2x^2+x-3) is x = 1

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domain of f(x)=sqrt(2x^2+x-3)

domain

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

x \leq - \frac{3}{2} or x \geq 1

Interval Notation

(- \infty, - \frac{3}{2}] \cup [1, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - \frac{3}{2} or x \geq 1

The function domain

x \leq - \frac{3}{2} or x \geq 1

2 4^{- x} + 8

\frac{x + 2}{x ^{2} - 16}

- 2 x^{3} - 6 x^{2} + 18 x + 1

f (x) = 11

f (y) = 2 x + b

What is the domain of f(x)=sqrt(2x^2+x-3) ?
The domain of f(x)=sqrt(2x^2+x-3) is x<=-3/2 \lor x>= 1