What is the domain of f(x)=(|x+1|-3)/(1+|x-3|) ?

The domain of f(x)=(|x+1|-3)/(1+|x-3|) is -infinity <x<infinity

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domain of f(x)=(|x+1|-3)/(1+|x-3|)

domain

f (x) = \frac{∣ x + 1 ∣ - 3}{1 + ∣ x - 3 ∣}

- \infty < x < \infty

Interval Notation

(- \infty, \infty)

Solution steps

Find undefined (singularity) points:

No real undefined points

The function has no undefined points nor domain constraints. Therefore, the domain is

- \infty < x < \infty

f (x) = \frac{1}{x + 5} + 8

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

y = lo g_{2} (4 - x^{2})

f (x) = \frac{3 x - 7}{5 x + 6}

\frac{x ^{2} - 8 x + 16}{3 x - 12}

What is the domain of f(x)=(|x+1|-3)/(1+|x-3|) ?
The domain of f(x)=(|x+1|-3)/(1+|x-3|) is -infinity <x<infinity