Question

Which of the following is true about the function below? (will draw in a sec!)

(and will also post answer choices in a sec!!)

***PLease explain!! :) thanks :D

Answer 1

this is the function :)

Answer 2

asnwers A,B,C,D (top to bottom)

Answer 3

Hi!! Thanks for coming!! :) DO u get this?? I am uber confused :(

Answer 4

you can't divide by zero, so \(\Large \sqrt{x-6} \neq 0\)

Answer 5

Well, it's domain is when \( \sqrt{x-6} \neq 0 \) (can't divide by 0) and \(x-6 \geq 0\) (no negative square roots).

It's range must be above 0, since it can't ever be negative.

Answer 6

If the denominator were zero, then

\(\Large \sqrt{x-6} = 0\)

\(\Large x-6 = 0^2\)

\(\Large x-6 = 0\)

\(\Large x = 0+6\)

\(\Large x = 6\)

so \(\Large x = 6\) makes the denominator zero, so \(\Large x \neq 6\) to avoid division by zero

Answer 7

so how do i find the range and domain tho? i don't get that :(

Answer 8

so x can't equal 6 and it can't equal sq. rt. of x-6 ?? did i get that right?

Answer 9

also, you can't take the square root of a negative number, so the radicand is positive (it's not zero for reasons stated above)

\(\Large x - 6 > 0\)

\(\Large x > 0+6\)

\(\Large x > 6\)

So the domain is \(\Large x > 6\) which in interval notation is \(\Large \left(6,\infty\right)\)

Answer 10

oh i get how to do that now! :) but how do i find the range?

Answer 11

to find the range, you start with

\[\Large y = \frac{1}{\sqrt{x-6}}\]

You then swap x and y

\[\Large x = \frac{1}{\sqrt{y-6}}\]

Then solve for y

\[\Large x\sqrt{y-6} = 1\]

\[\Large \sqrt{y-6} = \frac{1}{x}\]

\[\Large y-6 = \left(\frac{1}{x}\right)^2\]

\[\Large y-6 = \frac{1}{x^2}\]

\[\Large y = \frac{1}{x^2}+6\]

\[\Large y = \frac{1}{x^2}+6*\frac{x^2}{x^2}\]

\[\Large y = \frac{1}{x^2}+\frac{6x^2}{x^2}\]

\[\Large y = \frac{1+6x^2}{x^2}\]


Now find the domain of the last equation. This will give you the range of the original function.

Answer 12

y=7?

Answer 13

you can't divide by zero right?

Answer 14

no :/

Answer 15

but the only answer choices left that would work are A and C right? with parentheses right? not brackets??