How do you find domain and range for $f(x) =sqrt (x^2 - 2x + 5)$ ?

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Answer 1 · 2015-10-04T16:48:14

Complete the square to find that the domain of $f(x)$ is the whole of $RR$ and its range is $[2, oo)$

$f(x) = sqrt(x^2-2x+5) = sqrt((x-1)^2+4)$

$(x-1)^2+4 >= 4 > 0$ for all $x in RR$

So $f(x) = sqrt((x-1)^2+4)$ is defined for all $x in RR$

So the (implicit) domain of $f(x)$ is $RR$ .

$f(x)$ has minimum value when $(x-1) = 0$ , that is when $x = 1$ .

$f(1) = sqrt(0^2+4) = sqrt(4) = 2$

So the range of $f(x)$ is $[2, oo)$

How do you find domain and range for f(x) =sqrt (x^2 - 2x + 5)f(x) =sqrt (x^2 - 2x + 5)?