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

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

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

domain

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

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

Interval Notation

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

Solution steps

Find undefined (singularity) points:

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

The function domain

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

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

\frac{x ( x - 5 ) ^{2}}{4}

f (x) = \frac{1}{2} x^{2} - 5

y = (\frac{57}{23})^{- 6}

\frac{x - 4}{x - 1 + 2}

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