What is the range of f(x)=sqrt(x^2-25) ?

The range of f(x)=sqrt(x^2-25) is f(x)>= 0

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range of f(x)=sqrt(x^2-25)

range

f (x) = x^{2} - 25

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} - 25 : x \leq - 5 or x \geq 5

Extreme Points of

x^{2} - 25 :

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

f (x) = 3 x^{4} + 4 x^{3} + 6 x^{2} - 4

f (x) = 0.2 (x - 1)^{2} - 4

(04) (0 - 3)

f (x) = 3 x e^{2 x}

f (t) = 2.3 cos (0.25 t)

What is the range of f(x)=sqrt(x^2-25) ?
The range of f(x)=sqrt(x^2-25) is f(x)>= 0