What is the domain of f(x)=(sqrt(3-2x))/(x^2-64) ?

The domain of f(x)=(sqrt(3-2x))/(x^2-64) is x<-8\lor-8<x<= 3/2

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domain of f(x)=(sqrt(3-2x))/(x^2-64)

domain

f (x) = \frac{3 - 2 x}{x ^{2} - 64}

x < - 8 or - 8 < x \leq \frac{3}{2}

Interval Notation

(- \infty, - 8) \cup (- 8, \frac{3}{2}]

Solution steps

Find non-negative values for radicals:

x \leq \frac{3}{2}

Find undefined (singularity) points:

x = 8, x = - 8

Combine real regions and undefined points for final function domain

x < - 8 or - 8 < x \leq \frac{3}{2}

x^{2} + 6 x + 9

f (x) = (x - 1) (2 x + 3)

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

(- 5, 2), (4, 5)

f (x) = x^{2} + 4

What is the domain of f(x)=(sqrt(3-2x))/(x^2-64) ?
The domain of f(x)=(sqrt(3-2x))/(x^2-64) is x<-8\lor-8<x<= 3/2