What is the domain of f(x)=(\sqrt[4]{x^2-5x})/(x+1) ?

The domain of f(x)=(\sqrt[4]{x^2-5x})/(x+1) is x = 5

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domain of f(x)=(\sqrt[4]{x^2-5x})/(x+1)

domain

x < - 1 or - 1 < x \leq 0 or x \geq 5

Interval Notation

(- \infty, - 1) \cup (- 1, 0] \cup [5, \infty)

Solution steps

Find non-negative values for radicals:

x \leq 0 or x \geq 5

Find undefined (singularity) points:

x = - 1

Combine real regions and undefined points for final function domain

x < - 1 or - 1 < x \leq 0 or x \geq 5

y = \frac{x ^{2} - 3}{ln ( x ^{2} - 3 )}

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

3 + (x - 2)^{2}

ln ((x - 1)^{2})

f (x) = x^{2} + 3 x - 10

What is the domain of f(x)=(\sqrt[4]{x^2-5x})/(x+1) ?
The domain of f(x)=(\sqrt[4]{x^2-5x})/(x+1) is x<-1\lor-1<x<= 0\lor x>= 5