Answer:
[tex]x \ge 1[/tex]
Step-by-step explanation:
Given
[tex]\sqrt{6(x - 1)} \div \sqrt{3x^2[/tex]
Required
x for [tex]which[/tex] the function is [tex]de fi ned[/tex]
To do this, we set the numerator root [tex]\ge 0[/tex]
So, we have:
[tex]6(x - 1) \ge 0[/tex]
Divide by 6
[tex](x - 1) \ge 0[/tex]
Solve for x
[tex]x \ge 1[/tex]
Also, we set the denominator greater than 0:
[tex]3x^2 > 0[/tex]
Divide by 3
[tex]x^2 > 0[/tex]
Take square roots
[tex]x > 0[/tex]
So, we have:
[tex]x > 0[/tex] and [tex]x \ge 1[/tex]
Combine both to give:
[tex]x \ge 1[/tex]
