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

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

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

range

x^{2} - 2 x + 5

f (x) \geq 2

Solution steps

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

Domain of

x^{2} - 2 x + 5 : - \infty < x < \infty

Extreme Points of

x^{2} - 2 x + 5 :

Minimum

(1, 2)

Find the range for the interval

- \infty < x < \infty : 2 \leq f (x) < \infty

Combine the ranges of all domain intervals to obtain the function range

f (x) \geq 2

x = 5

f (x) = - \frac{6}{x - 7}

y = \frac{1}{2} cos (3 x + π) - 5

f (x) = \frac{x ^{2} + 20 x + 1}{x ^{2} + 2 x + 1}

f (x) = - \frac{1}{x ^{2} + 5}