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

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

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domain of y= 9/((2x+1)^2)

domain

y = \frac{9}{( 2 x + 1 ) ^{2}}

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}

y = x + 2 cos (x)

y = - \frac{5}{2} x + 2

f (t) = 4 sin (t)

y = cot (x - \frac{π}{4})

f (x) = x^{3} - 5 x^{2} + x - 5

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