1 Answer

Ratzo · Answer 1 · 2022-12-03T06:01:44+0000

Answer:

$x \ge 1$

Explanation:

Given

$√(6(x - 1)) / \sqrt{3x^2$

Required

x for
which the function is
de fi ned

To do this, we set the numerator root
$\ge 0$

So, we have:

$6(x - 1) \ge 0$

Divide by 6

$(x - 1) \ge 0$

Solve for x

$x \ge 1$

Also, we set the denominator greater than 0:

3x^2 > 0

Divide by 3

x^2 > 0

Take square roots

x > 0

So, we have:

x > 0 and
$x \ge 1$

Combine both to give:

$x \ge 1$

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