What is the domain of #h(x)=sqrt(x^2 - 2x + 5)#?

1 Answer
Stefan V.
Sep 6, 2015

Domain: #(-oo, + oo)#

Explanation:

Since you're dealing with the square root of an expression, you know that you need to exclude from the domain of the function any value of #x# that will make the expression under the square root negative.

For real numbers, the square root can only be taken from positive numbers, which means that you need

#x^2 - 2x + 5 >=0#

Now you need to find the values of #x# for which the above inequality is satisfied. Look what happens when you use a little algebraic manipulation to rewrite the inequality

#x^2 - 2x + 5 >= 0#

#x^2 - 2x + 1 + 4 >=0#

#(x-1)^2 + 4 >=0#

Because #(x-1)^2 >=0# for any value of #x in RR#, it follows that

#(x-1)^2 + 4 >=0", "(AA)x in RR#

This means that the domain of the function can include all real numbers, since you cannot have a negative expression under the square root regardless of which #x# you plug in.

In interval notation, the domain of the function will thus be #(-oo, + oo)#.

graph{sqrt(x^2-2x+5) [-10, 10, -5, 5]}

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