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

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

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range of f(x)=sqrt(x^2-5x)

range

f (x) = 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

x^{3} - 3 x

f (x) = - tan (x + 6) - 5

f (x) = \frac{8}{\frac{3}{x} - 1}

(- 5, 5), (2, - 5)

3 x 3 x^{2} + 4

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