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

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

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

range

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

f (x) \geq 2

Interval Notation

[2, \infty)

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

f (x) = \frac{3 x}{6 x - 5}

f (x) = - x^{3} - 9

f (x) = \frac{5}{x ^{2} + 4 x}

f (x) = \frac{7 x + 2 x}{x}

f (x) = \frac{x ^{2} + 2 x + 4}{x ^{2} + 2 x}

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