What is the domain of (16-t)^{1/8} ?

The domain of (16-t)^{1/8} is t<= 16

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domain of (16-t)^{1/8}

domain

(16 - t)^{\frac{1}{8}}

t \leq 16

Interval Notation

(- \infty, 16]

Solution steps

Find non-negative values for radicals:

t \leq 16

The function domain

t \leq 16

x^{2} - 4 x + 3

f (x) = \frac{x + 2}{x - 6}

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

3 x - 8

y = \frac{1}{x + 1}