What is the range of sqrt(x^2-1) ?

The range of sqrt(x^2-1) is f(x)>= 0

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range of sqrt(x^2-1)

range

x^{2} - 1

f (x) \geq 0

Interval Notation

[0, \infty)

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

x^{2} - 1 : x \leq - 1 or x \geq 1

Extreme Points of

x^{2} - 1 :

Minimum

(- 1, 0),

Minimum

(1, 0)

Find the range for the interval

- \infty < x \leq - 1 : 0 \leq f (x) < \infty

Find the range for the interval

1 \leq x < \infty : 0 \leq f (x) < \infty

Combine the ranges of all domain intervals to obtain the function range

f (x) \geq 0 or f (x) \geq 0

f (x) \geq 0

\frac{x ^{2} - 9}{x ^{2} + 1}

x^{2} - x

(- 2, 4), (- 6, - 1)

x^{2} - 4 x

f (x) = - 2 x^{3} - 24 x^{2} - 72 x