Which of the following is true about the function below? 1/√ x-6 A. Its domain is [6,∞) and its range is . (-∞,0) U (0,∞) B. Its domain is (6,∞) and its range is . (0,∞) C. Its domain is (6,∞) and its range is . (-∞,∞) D. Its domain is (-∞,0] and its range is . (0,∞)

Question

anonymous · Answer

The domain will be all numbers that prevent the denominator from becoming zero, or otherwise undefined.

Where is $\sqrt{x-6}$ defined? In other words, finding the domain of this will give the domain of $\dfrac{1}{\sqrt{x-6}}$.

As for range, consider what kind of values can't be attained. Obviously, you can get negative numbers because the square root function never yields a negative number, so you can eliminate that part of the potential range, $(-\infty,0)$. What about 0? Can the fraction ever attain 0? What about the positive numbers?

anonymous · Answer

so everyone that has a zero is wrong?

anonymous · Answer

I can't draw an answer from what you gave me.

anonymous · Answer

For the fraction to be defined, you must have $x>6$. If $x=6$, you get $\dfrac{1}{\sqrt0}$, which is undefined. If $x<6$, you get $\dfrac{1}{\sqrt{\text{negative number}}}$, also undefined. So, the domain is all numbers strictly greater than 6, or $(6,\infty)$.

anonymous · Answer

For the range, I suggested you think about what kind of values the fraction can't possibly attain. 

Can $\dfrac{1}{\sqrt{\cdots}}$ ever be negative? (No, the square root only yields positive values.)

Can $\dfrac{1}{\sqrt{\cdots}}$ ever be zero? (No, this is not possible for any fraction so long as the numerator is non-zero. Think about $\dfrac{1}{x}$, which is undefined at 0, as an example.)

This means the range must be $(0,\infty)$, since nothing is barring the fraction from taking on any positive number.