What is the domain of f(x)=(15)/(16x^2-1) ?

The domain of f(x)=(15)/(16x^2-1) is x 1/4

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domain of f(x)=(15)/(16x^2-1)

domain

f (x) = \frac{15}{16 x ^{2} - 1}

x < - \frac{1}{4} or - \frac{1}{4} < x < \frac{1}{4} or x > \frac{1}{4}

Interval Notation

(- \infty, - \frac{1}{4}) \cup (- \frac{1}{4}, \frac{1}{4}) \cup (\frac{1}{4}, \infty)

Solution steps

Find undefined (singularity) points:

x = \frac{1}{4}, x = - \frac{1}{4}

The function domain

x < - \frac{1}{4} or - \frac{1}{4} < x < \frac{1}{4} or x > \frac{1}{4}

f (x) = \frac{x}{x - 16}

f (x) = \frac{x - 3}{5 x - 2}

f (x) = \frac{1}{36 - x}

- 2 x^{2} + 16 x - 31

lo g_{10} (4 - x)

What is the domain of f(x)=(15)/(16x^2-1) ?
The domain of f(x)=(15)/(16x^2-1) is x<-1/4 \lor-1/4 <x< 1/4 \lor x> 1/4