What is the range of f(x)=sqrt(x^2-3x) ?

The range of f(x)=sqrt(x^2-3x) is f(x)>= 0

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range of f(x)=sqrt(x^2-3x)

range

f (x) = x^{2} - 3 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} - 3 x : x \leq 0 or x \geq 3

Extreme Points of

x^{2} - 3 x :

Minimum

(0, 0),

Minimum

(3, 0)

Find the range for the interval

- \infty < x \leq 0 : 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

f (x) = cot (x)

f (x) = \frac{x - 3}{2}

y = tan (x)

x^{2} + 4

(0, - 1), (8, 7)

What is the range of f(x)=sqrt(x^2-3x) ?
The range of f(x)=sqrt(x^2-3x) is f(x)>= 0