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

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

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

domain

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

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

y = lo g_{2} (4 - x^{2})

f (x) = \frac{3 x - 7}{5 x + 6}

\frac{x ^{2} - 8 x + 16}{3 x - 12}

f (x) = \frac{x - 3}{x ^{2} - 4 x - 32}

(x^{2} + 2 x - 1) (2 x^{2} - 3 x + 6)

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