What is the range of sqrt(x^2-5x+6) ?

The range of sqrt(x^2-5x+6) is f(x)>= 0

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range of sqrt(x^2-5x+6)

range

x^{2} - 5 x + 6

f (x) \geq 0

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

x^{2} - 5 x + 6 : x \leq 2 or x \geq 3

Extreme Points of

x^{2} - 5 x + 6 :

Minimum

(2, 0),

Minimum

(3, 0)

Find the range for the interval

- \infty < x \leq 2 : 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

x^{2} - 3 x - 4

f (x) = 7 x^{3} - 4

f (x) = \frac{x - 1}{x ^{2} - x}