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

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

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

domain

f (x) = \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) = cosh (t^{3} - 8)

f (x) = \frac{14 - 5 x}{3 x + 4}

g (x) = lo g_{2} (7 - x)

- lo g_{2} (x - 1) + 3

f (x) = x^{4} - 3 x^{3} + 7 x^{2} - 5 x + 2

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