The domain is simply stated as the x-values. The domains for a linear, diagonal line is all real numbers. On the other hand, the domain can be limited by many ways.
ex.) f(x) = log(x)
That means ~>
Does 10 to any power give you a negative number or zero? Let's try:
10^2 = 10*10 = 100
10^0 = 1
10^(-2) = 1/10^2 = 1/(10*10) = 1/100
No, ten to any power does not give you a negative number or zero.
Domain: { x | }
ex.) f(x) = sqrt(x - 1)
Can you take the square root of a negative number? No. Can you take the square root of zero? Yes.
So, what is in the square root must be equal to or greater than zero.
Domain: { x | x >= 1 }
ex.) f(x) = 1/sqrt(x + 3)
Can you divide by zero? Yes, but it would be wrong. Your answer must be greater to or equal to due to the square root. Your answer must be greater than zero due to the square root in the denominator.
Domain: { x | x > -3 }
ex.) f(x) = log(x + 1)/sqrt(3 - x)
Log of a number must be greater than zero: x + 1 > 0 which is x > -1
Square root of a number in the denominator must be greater than zero: 3 - x > 0 which is x < 3
So you have:
<--- (-1) ~~~ (3) ~~~>
<~~~ (-1) ~~~ (3) --->
Result: <--- (-1) ~~~ (3) --->
Domain: { x | -1 < x < 3 }
ex.) Which has a greater expanse of domain? f(x) = log(sqrt(x)) ~or~ f(x) = sqrt(log(x))
f(x) = log(sqrt(x))
sqrt(x) ~>
log(x) ~>
Domain: { x | x > 0 } More Expansive
f(x) = sqrt(log(x))
The log(.01) = -2
sqrt(-1) = i Does not work
The log(1) = 0
sqrt(0) = 0
Domain: { x | x >= 1 }