What is the domain of (6x^2+7x-3)/(3x^2-4x+1) ?

The domain of (6x^2+7x-3)/(3x^2-4x+1) is x 1

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domain of (6x^2+7x-3)/(3x^2-4x+1)

domain

\frac{6 x ^{2} + 7 x - 3}{3 x ^{2} - 4 x + 1}

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

Interval Notation

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

Solution steps

Find undefined (singularity) points:

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

The function domain

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

f (x) = \frac{1}{5 x - 15}

x^{2}, x < 0

f (x) = 6 x - x^{2}

f (x) = lo g_{10} (7 - 4 x)

ln (2 x)

What is the domain of (6x^2+7x-3)/(3x^2-4x+1) ?
The domain of (6x^2+7x-3)/(3x^2-4x+1) is x< 1/3 \lor 1/3 <x<1\lor x>1