What is the domain of (12x^2+24)/(3x^2-4x-7) ?

The domain of (12x^2+24)/(3x^2-4x-7) is x 7/3

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domain of (12x^2+24)/(3x^2-4x-7)

domain

\frac{12 x ^{2} + 24}{3 x ^{2} - 4 x - 7}

x < - 1 or - 1 < x < \frac{7}{3} or x > \frac{7}{3}

Interval Notation

(- \infty, - 1) \cup (- 1, \frac{7}{3}) \cup (\frac{7}{3}, \infty)

Solution steps

Find undefined (singularity) points:

x = \frac{7}{3}, x = - 1

The function domain

x < - 1 or - 1 < x < \frac{7}{3} or x > \frac{7}{3}

f (x) = \frac{4 x}{x ^{2} + 2 x - 15}

3 x^{3} + 4 x^{2} - x - 4

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

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

y + 5 = - x + 1

What is the domain of (12x^2+24)/(3x^2-4x-7) ?
The domain of (12x^2+24)/(3x^2-4x-7) is x<-1\lor-1<x< 7/3 \lor x> 7/3