What is the range of sqrt(x^2-5) ?

The range of sqrt(x^2-5) is f(x)>= 0

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range of sqrt(x^2-5)

range

x^{2} - 5

f (x) \geq 0

Solution steps

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

Domain of

x^{2} - 5 : x \leq - 5 or x \geq 5

Extreme Points of

x^{2} - 5 :

Minimum

(- 5, 0),

Minimum

(5, 0)

Find the range for the interval

- \infty < x \leq - 5 : 0 \leq f (x) < \infty

Find the range for the interval

5 \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

2 sin (4 x - π)

(4.5) (6.7)

4 x + 3 y = 12

f (x) = 6 (\frac{x}{2}) - 5

f (x) = x^{3} + 15 x^{2}