calculus domain question?

Question

f(x)= ____1______
sqrt(15x+11)

I need the domain of the problem please. In English, it reads...

1 divided by the square root of 15X plus 11.

Thank you in advance!

Answer 1

The domain is the values that you are allowed to use for x. In this case, you are not allowed to divide by zero, and you are not allowed to take the square root of a negative number. So 15x + 11 must be greater than 0.

15x + 11 > 0
15x > -11
x > -11/15

So your domain is { x | x > -11/15 }.

"But Louis! You said we couldn't use negative numbers!"

No I didn't. I said we cannot take the square root of a negative number. This function takes the square root of 15x + 11, so 15x + 11 must not be negative. 15x + 11 must also not be zero to avoid division by zero. As long as x > -11/15, then 15x + 11 > 0.

I hope this helps! *tips hat to MathGirl as he leaves*

Answer 2

Ok you have to learn this stuff yourself and not have others do your homework for you.

The domain is the x values for which your equation is valid. So you know first of all that there cannot be a 0 on the bottom. You also have sqrt(15x+11). You also know that you cannot have a negative square root.

So now you have two conditions- sqrt(15x+11) cannot equal 0 and that x has to be a number such that 15x+11 is greater than 0.

Now you can solve for x- 15x>-11 so x> -11/15

Answer 3

The area are the values of x for which f(x) exists. the decision are the values f(x) can take a million) f(x) = x^2 + 4 there is not any value of x which promises you with hassle. x^2 is continuously more effective than or equivalent to 0 so f(x)>=4 on the notation round brackets ( are open periods and sq. brackets [ are closed periods. decision [4,+?) 2)f(x)= ?(x^2 + 5x + 6 sq. roots of adverse numbers do no longer exist (in the set of authentic numbers) (x^2 + 5x + 6) = (x+2)(x+3) is decrease than 0 if -3

Answer 4

15x + 11 > 0
15x > -11
x > -11/15

Answer 5

Darn, to late. Mathgirl did it again.