What is the domain of f(x)= 1/(sqrt(2x^2+3x-54)) ?

The domain of f(x)= 1/(sqrt(2x^2+3x-54)) is x 9/2

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domain of f(x)= 1/(sqrt(2x^2+3x-54))

domain

f (x) = \frac{1}{2 x ^{2} + 3 x - 54}

x < - 6 or x > \frac{9}{2}

Interval Notation

(- \infty, - 6) \cup (\frac{9}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - 6 or x \geq \frac{9}{2}

Find undefined (singularity) points:

x = \frac{9}{2}, x = - 6

Combine real regions and undefined points for final function domain

x < - 6 or x > \frac{9}{2}

\frac{( x + 1 ) ^{3}}{( x - 1 ) ^{2}}

f (x) = lo g_{e} (x - 5)

p (x) = 3 x + 12

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

∣ x ∣ + x + 2

What is the domain of f(x)= 1/(sqrt(2x^2+3x-54)) ?
The domain of f(x)= 1/(sqrt(2x^2+3x-54)) is x<-6\lor x> 9/2