What is the domain of f(x)=(x+1)/(2x^2+7x-9) ?

The domain of f(x)=(x+1)/(2x^2+7x-9) is x 1

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domain of f(x)=(x+1)/(2x^2+7x-9)

domain

f (x) = \frac{x + 1}{2 x ^{2} + 7 x - 9}

x < - \frac{9}{2} or - \frac{9}{2} < x < 1 or x > 1

Interval Notation

(- \infty, - \frac{9}{2}) \cup (- \frac{9}{2}, 1) \cup (1, \infty)

Solution steps

Find undefined (singularity) points:

x = 1, x = - \frac{9}{2}

The function domain

x < - \frac{9}{2} or - \frac{9}{2} < x < 1 or x > 1

f (x) = \frac{2}{2 x + 6}

f (x) = \frac{1}{7 x + 4}

f (x) = - 5 x^{2} + 85 x + 9

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

f (x) = 2 x - \frac{6}{5}

What is the domain of f(x)=(x+1)/(2x^2+7x-9) ?
The domain of f(x)=(x+1)/(2x^2+7x-9) is x<-9/2 \lor-9/2 <x<1\lor x>1