I need a refresher on finding range & domain please!
f(x)= 1+x2
What is the range & domain?
Domain: If the expression does NOT contain either of these:
(1) denominators containing variables
(2) radicals of EVEN index (square root, 4th root, 6th root,...)
then the domain in always the same
in set-builder notation: {x|x is a real number}
in interval notation (¥,¥)
If a denominator contains variables:
(1) find the zeros of any denominator
(2) if there are no real zeros, then domain is as above
in set-builder notation: {x|x is real}
in interval notation (-¥,¥)
(3) If the denominator has zeros, say, a, b,...(etc.)...z.
from smallest to largest
then the domain is
in set-builder notation:
{x|a < x < b,c < x < d,...etc.,y < x < z}
in interval notation: {-¥,a) È (a,b) È (b,c)...etc. È (z,¥)
If there is an radical with even index:
(1) If domain is in the numerator, solve the inequality
NUMERATOR ³ 0 (greater than or equal to)
(2) If domain is in the denominator, solve the inequality
NUMERATOR > 0 (strictly greater than)
(3) Domain is the solution to that inequality.
To find the range, replace f(x) by y and solve for x.
The rules for the range are the same as the rules for the domain
except that every "x" is replaced by a y.
Now for your problem
f(x)= 1+x2
It doesn't contain a denominator or an even-index radical, so the domain
is simply {x|x is a real number} in set-builder notation or (-¥, ¥) in
interval notation.
For the range:
f(x) = 1 + x2
Replace f(x) by y
y = 1 + x2
Solve for x
x2 = y - 1
_____
x = ±Öy - 1
That is a radical with an even index, so we solve the inequality
y - 1 ³ 0
y ³ 1
In set builder notation range = {y|y ³ 1}
In interval notation range = [1, ¥)
Edwin
AnlytcPhil@aol.com
