What is the domain of ((n-1)(-1)^{n+1})/(2n+3) ?

The domain of ((n-1)(-1)^{n+1})/(2n+3) is n -3/2

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domain of ((n-1)(-1)^{n+1})/(2n+3)

domain

\frac{( n - 1 ) ( - 1 ) ^{n + 1}}{2 n + 3}

n < - \frac{3}{2} or n > - \frac{3}{2}

Interval Notation

(- \infty, - \frac{3}{2}) \cup (- \frac{3}{2}, \infty)

Solution steps

Find undefined (singularity) points:

n = - \frac{3}{2}

The function domain

n < - \frac{3}{2} or n > - \frac{3}{2}

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

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

f (x) = - 16 t^{2} + 64 t + 160

f (x) = - 6 x + 4

\frac{2 x + 4}{x ^{2} + 3 x + 2}

What is the domain of ((n-1)(-1)^{n+1})/(2n+3) ?
The domain of ((n-1)(-1)^{n+1})/(2n+3) is n<-3/2 \lor n>-3/2