What is the domain of g(x)=sqrt(4-x)+sqrt(x^2-1) ?

The domain of g(x)=sqrt(4-x)+sqrt(x^2-1) is x<=-1\lor 1<= x<= 4

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domain of g(x)=sqrt(4-x)+sqrt(x^2-1)

domain

g (x) = 4 - x + x^{2} - 1

x \leq - 1 or 1 \leq x \leq 4

Interval Notation

(- \infty, - 1] \cup [1, 4]

Solution steps

Find non-negative values for radicals:

x \leq - 1 or 1 \leq x \leq 4

The function domain

x \leq - 1 or 1 \leq x \leq 4

f (x) = ∣ x + 2 ∣ + ∣ x - 2 ∣

y = 3 x + 2

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

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

y = \frac{x ^{2} - 81}{x - 9}

What is the domain of g(x)=sqrt(4-x)+sqrt(x^2-1) ?
The domain of g(x)=sqrt(4-x)+sqrt(x^2-1) is x<=-1\lor 1<= x<= 4