domains and functions

[wildcornstalk]

Find the domain of the function and express the answer in interval notation. Explain in words or show the calculations.
A. F(x)=3x-1 (-infinity, infinty)
B. G(x)=sqrt(x+5) (-infinity, -5)
C. F(x)=16x/(x^2+9) (-infinity,-3)U(-3,3)U(3, infinity)
D. G(x)=13x^2-5x+9 (-infinity, infinity)
E. F(x)=6/(x-5) (-infinity,5)U(5, infinity)


Am I doing this correctly? I am doubting my answers.Especially c, do I take the square root of 9?

 

[mmm4444bot]

Where are you stuck? I mean, what have you thought about or done thus far?

Please be specific.

If you're not sure about the word domain, I would start by looking at the index of your textbook.

Cheers

 

[twlectures]

c.) Your answer is: F(x)=16x/(x^2+9) (-infinity,-3)U(-3,3)U(3, infinity)

I would say that x can take on any value from -inf to +inf.

F(x) = 0 when x -> -inf
F(x) = 0 when x -> 0
F(x) = 0 when x -> +inf
F(x) = 16/10 when x = 1
F(x) = -16/10 when x = -1
...

There is NO value of x that will give a value of 0 in the denominator since it is x^2 + 9 (not true if it is x^2 - 9)
(This is of course ignoring imaginary numbers which I think we can do at this level in the game)

So, I would say that the Domain of the function is between -inf and +inf.

 

[twlectures]

Your Answer: G(x)=sqrt(x+5) (-infinity, -5)


Seems like the domain should be (-5, +infinity) since you want the argument of the sqrt() to always be positive.

For your answer, G(-10) = sqrt( -10 + 5 ) = sqrt( -5 ) --> Not allowed at this point in the Math game.

Regards,

 

[mmm4444bot]

Am I doing this correctly?

You seem to understand the general concept of domain.

Of course, the instructions require you to also include either an explanation in words or a listing of your steps, to demonstrate how you arrived at your answers.

You've made some mistakes with the numbers, but your interval notations look proper.


A. F(x) = 3x - 1 (-infinity, infinty)

D. G(x) = 13x^2 - 5x + 9 (-infinity, infinity)

These two are easy; the domain of all polynomial functions is the set of Real numbers.


B. G(x) = sqrt(x + 5) (-infinity, -5)

Somebody already explained why this is incorrect, but they overlooked the fact that the radicand may be zero.

Did you solve the inequality x + 5 ? 0 ?


C. F(x) = 16x/(x^2 + 9) (-infinity,-3) U (-3,3) U (3, infinity)

do I take the square root of 9? No

I think that you made the same mistake here as you made in your other exercise, where the denominator is the same sum of squares.

When solving the equation x^2 + 9 = 0, you first subtract 9 from both sides. When you see x^2 = -9, you should recognize that no square is negative and conclude that the equation has no Real solutions. Hence, there are no values that will lead to zero in the denominator of function F.


E. F(x) = 6/(x - 5) (-infinity,5)U(5, infinity)

Correct!

 

[mmm4444bot]

G(x) = sqrt(x + 5)

Seems like the domain should be (-5, +infinity) since you want the argument of the sqrt() to always be positive.

Your phrase in red above is incorrect.

The argument of the sqrt() function is always "non-negative", not "positive".

In other words, we can take the square root of zero.

This means that the value x = -5 is included in the domain of function G, and we must change your interval notation above accordingly.

Cheers