What is the domain of f(x)=((x+1)^{1/3})/(x^2-5x+4) ?

The domain of f(x)=((x+1)^{1/3})/(x^2-5x+4) is x 4

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domain of f(x)=((x+1)^{1/3})/(x^2-5x+4)

domain

f (x) = \frac{( x + 1 ) ^{\frac{1}{3}}}{x ^{2} - 5 x + 4}

x < 1 or 1 < x < 4 or x > 4

Interval Notation

(- \infty, 1) \cup (1, 4) \cup (4, \infty)

Solution steps

Find undefined (singularity) points:

x = 4, x = 1

The function domain

x < 1 or 1 < x < 4 or x > 4

- 6 \cdot 3 \cdot x + 6 + 2 \cdot 3 \cdot π

f (x) = - \frac{3}{2} tan (x - \frac{π}{6})

(- 3.5) (4.4)

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

f (x) = 4 - x^{2}

What is the domain of f(x)=((x+1)^{1/3})/(x^2-5x+4) ?
The domain of f(x)=((x+1)^{1/3})/(x^2-5x+4) is x<1\lor 1<x<4\lor x>4