What is the domain of f(x)= 1/(\sqrt[4]{x^2-2x-8)} ?

The domain of f(x)= 1/(\sqrt[4]{x^2-2x-8)} is x 4

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domain of f(x)= 1/(\sqrt[4]{x^2-2x-8)}

domain

f (x) = \frac{1}{4 x ^{2} - 2 x - 8}

x < - 2 or x > 4

Interval Notation

(- \infty, - 2) \cup (4, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - 2 or x \geq 4

Find undefined (singularity) points:

x = 4, x = - 2

Combine real regions and undefined points for final function domain

x < - 2 or x > 4

f (x) = \frac{1 + 3 ( 1 ^{2} - 1 )}{2}

4 + \frac{1}{x}

x^{2} - 16 x + 66

\frac{x + 6}{12 - x ^{2} - 25}

f (x) = \frac{2 x + 3}{5 x - 2}

What is the domain of f(x)= 1/(\sqrt[4]{x^2-2x-8)} ?
The domain of f(x)= 1/(\sqrt[4]{x^2-2x-8)} is x<-2\lor x>4