What is the domain of f(x)= 1/(x^2+x-6)+sqrt(-x+1) ?

The domain of f(x)= 1/(x^2+x-6)+sqrt(-x+1) is x<-3\lor-3<x<= 1

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domain of f(x)= 1/(x^2+x-6)+sqrt(-x+1)

domain

f (x) = \frac{1}{x ^{2} + x - 6} + - x + 1

x < - 3 or - 3 < x \leq 1

Interval Notation

(- \infty, - 3) \cup (- 3, 1]

Solution steps

Find non-negative values for radicals:

x \leq 1

Find undefined (singularity) points:

x = 2, x = - 3

Combine real regions and undefined points for final function domain

x < - 3 or - 3 < x \leq 1

f (x) = ln (x^{2} + 64)

\frac{3 x}{8 x - 1}

y = \frac{7}{x ( x - 4 )}

y = \frac{6 x - 2 - 3}{x ^{2} + 1}

ln (4 - 3 x - x^{2})

What is the domain of f(x)= 1/(x^2+x-6)+sqrt(-x+1) ?
The domain of f(x)= 1/(x^2+x-6)+sqrt(-x+1) is x<-3\lor-3<x<= 1