Express the domain of
in interval notation.
Identify the only restriction. The numerator is a polynomial, defined everywhere; the function fails only where the denominator is zero. So the task reduces to solving .
Solve the biquadratic with a substitution. Only even powers appear, so put :
Return to x. Each positive gives two real values of :
Four excluded points in total — the maximum a quartic denominator can produce.
Remove the four points and write the intervals. Ordered on the number line, the excluded values are , which cut into five pieces:
Check one point in each interval. At : denominator . At : . At : . At : . At : . None vanishes, confirming the five intervals are entirely inside the domain.
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