How to find domain? Help!?

Question

find domain given the function f

f(x) = sqr(25 - x^2)

f(x) = sqr(x^2 - 5x)

f(x) = sqr((3 - x) / (x + 2))

can any one show me steps / methods plz
thanks

Answer 1

I'm just going to do one because I have the feeling that after I explain it, you'll be able to do the rest yourself. :)

f(x) = √(25 - x^2)

To find the domain, just think "what can the value of x be?" In a situation like this, x can really be almost anything, right? Let's look at a few examples:

Can x be 1?

f(x) = √(25 - 1^2) = √(25-1) = √24

Yep.

Can x be 2?

f(x) = √(25 - 2^2) = √(25-4) = √21

Sure!

Can x even be a negative number, like, for instance, -3?

f(x) = √(25 - [-3^2]) = √(25- [9]) = √16 = 4

Yes, it can!

However, now we start getting into a tricky situation... What about if x is a large number, such as 10?


f(x) = √(25 - 10^2) = √(25-100) = √-75

Uh-oh! We can't take the square root of a negative number, can we? (Well, technically later on we'll be able to by using the imaginary number, i, but for all intents and purposes at this point, we'll say that we can't.)

That means that 10 is NOT in the domain of x. In simple terms, it means that x cannot equal 10 because when it does, we don't get a "real" number... we get the square root of a negative number which is NOT real. Through some more trail and error we can see that any time x is 6 or larger, we will get a negative number under the square root sign.

All together this suggests that any time x is greater than or equal to 6, we won't have a real number. THEREFORE, x cannot be ≥ 6 or ≤ -6. (The reason why x can't be ≤ -6 is because when you take a square a negative number, you get a positive, so it's the equivalent to the positive version of it when dealing with squares.)

Since the domain is asking for what x CAN be and not what it CAN'T be, we simply spin that around and say that the domain of x is x ≤ 5. In fancy math terms, {x | -5 ≤ x ≤ 5}.

There you go!!! I hope that helped!

Answer 2

f(x) = sqr(25 - x^2)
lets try to substitute 5
= sqr(25 - (5)^2
= sqr(0)
= 0

lets try tlets try to substitute -5
= sqr(25 - (-5)^2
= sqr(0)
= 0

substitute 6
= sqr(25 - (6)^2)
= sqr(25 - 36)
= sqr(-11)
= undefine
therefore the domain is [-5, 5]

thats how to do it, just find a digit that could make the expression to zero and a digit that could make it undefine. In ex.1 if you try to put any digit greater than 5 or less than -5 the expression would become undefine. So the domain is only [-5, 5].

When I say [-5 ,5] it includes the no's -5,-4,-3,-2,-1,0,1,2,3,4,5

Answer 3

There is no "set" method to do this. You just have to "explore" the function to see where it exists and where it doesn't. This is why you are taught to sketch graphs.
1. This is the equation of a circle, with radius 5 centered at the origin. So the domain will be from -5 to +5 including these points.
2. This is the sqrt of a parabola which has a minimum at some point on the x-axis with a y-value of zero. Thus, at all points, a legitimate domain can exist.
3. You should be able to get the hang of these by now. Try it. Remember, if the arguement is negative or x=-2, you can't have a domain.

Answer 4

The first two functions have no finite poles, so have domain over all complex numbers, although the values for the first function for |x| > 5 are complex. The third function has a pole at x = -2, so the domain includes all real numbers other than that one. Again, the function has complex values for real x > 3.

Answer 5

domain is anything that doesn't violate the rules.

any number that can go into these equations, so if you find a rule breaker , such as 0 as denominator, negative in sqr, those are out of the domain!

Answer 6

-5 =< x =< 5 All other values of x make f(x) imaginary

-infinity
-2

Answer 7

Also make sure you are not taking the square root of a negative number.

1) -5<=x<=5