What is the range of sqrt(x^2-9) ?

The range of sqrt(x^2-9) is f(x)>= 0

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range of sqrt(x^2-9)

range

x^{2} - 9

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} - 9 : x \leq - 3 or x \geq 3

Extreme Points of

x^{2} - 9 :

Minimum

(- 3, 0),

Minimum

(3, 0)

Find the range for the interval

- \infty < x \leq - 3 : 0 \leq f (x) < \infty

Find the range for the interval

3 \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

m = \frac{5}{4}, (- 8, - 3)

f (x) = x^{4} - 32 x^{2} + 9

3 x - 2 y = - 6

f (x) = \frac{y}{y - 6} + \frac{15}{y + 6}

f (x) = \frac{- 15}{x ^{2} + 5}