What is the domain of f(x)=(sqrt(2-x))/(x^2+4x) ?

The domain of f(x)=(sqrt(2-x))/(x^2+4x) is x<-4\lor-4<x<0\lor 0<x<= 2

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domain of f(x)=(sqrt(2-x))/(x^2+4x)

domain

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

x < - 4 or - 4 < x < 0 or 0 < x \leq 2

Interval Notation

(- \infty, - 4) \cup (- 4, 0) \cup (0, 2]

Solution steps

Find non-negative values for radicals:

x \leq 2

Find undefined (singularity) points:

x = 0, x = - 4

Combine real regions and undefined points for final function domain

x < - 4 or - 4 < x < 0 or 0 < x \leq 2

f (x) = ln (x)

f (x) = 6 cos (3 x + \frac{π}{2})

f (x) = 0.55 \cdot x - 2.8 e^{5}

f (x) = 2^{\frac{x}{2}}

(0, 4), (- 3, 0)

What is the domain of f(x)=(sqrt(2-x))/(x^2+4x) ?
The domain of f(x)=(sqrt(2-x))/(x^2+4x) is x<-4\lor-4<x<0\lor 0<x<= 2