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

The domain of f(x)=(2x-3)/(4x^2+3x-1) is x 1/4

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domain of f(x)=(2x-3)/(4x^2+3x-1)

domain

f (x) = \frac{2 x - 3}{4 x ^{2} + 3 x - 1}

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

Interval Notation

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

Solution steps

Find undefined (singularity) points:

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

The function domain

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

f (x) = x^{2} - 4 x + 3

f (x) = x - 3 y = 6

- 1 - \frac{7}{x ^{2} - 9}

y = x^{4} + x^{2} - 4 x + 4

f (x) = \frac{x + 7}{x + 1}

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