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

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

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range of sqrt(x^2-5x)

range

x^{2} - 5 x

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} - 5 x : x \leq 0 or x \geq 5

Extreme Points of

x^{2} - 5 x :

Minimum

(0, 0),

Minimum

(5, 0)

Find the range for the interval

- \infty < x \leq 0 : 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) = 9 x + 48

y = x^{4} - x^{2} + 6

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

f (l) = l

f (x) = \frac{7}{x - 4}