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

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

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

domain

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

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

Interval Notation

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

Solution steps

Find undefined (singularity) points:

x = \frac{1}{2}

The function domain

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

f (x) = 1 + \frac{2}{x}, x \neq = 0

f (x) = \frac{1}{x ^{2} - 2 x - 35}

(\frac{3}{4})^{x}

y = 3 x - 7

\frac{x - 4}{x}

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