Question 24387
<pre>I need a refresher on finding range & domain please!
     f(x)= 1+x<sup>2</sup> 
What is the range & domain?
<font size = 4><b>
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 (<font face = "symbol">¥</font>,<font face = "symbol">¥</font>)

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 (-<font face = "symbol">¥</font>,<font face = "symbol">¥</font>)
        (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: {-<font face = "symbol">¥</font>,a) <font face = "symbol">È</font> (a,b) <font face = "symbol">È</font> (b,c)...etc. <font face = "symbol">È</font> (z,<font face = "symbol">¥</font>)

If there is an radical with even index:
        (1) If domain is in the numerator, solve the inequality
            NUMERATOR <font face = "symbol">³</font> 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+x<sup>2</sup>

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 (-<font face = "symbol">¥</font>, <font face = "symbol">¥</font>) in
interval notation.

For the range:

     f(x) = 1 + x<sup>2</sup>

Replace f(x) by y

        y = 1 + x<sup>2</sup>

Solve for x

      x<sup>2</sup> = y - 1
             _____
       x = ±<font face = "symbol">Ö</font>y - 1  

That is a radical with an even index, so we solve the inequality

              y - 1 <font face = "symbol">³</font> 0
                  
                  y <font face = "symbol">³</font> 1

In set builder notation range = {y|y <font face = "symbol">³</font> 1}
In interval notation range = [1, <font face = "symbol">¥</font>)

Edwin
AnlytcPhil@aol.com</pre>