What is the range of the function #f(x)=x^2-1#?

1 Answer
Apr 2, 2017

Answer:

#{y|y≥-1}#

Explanation:

Basically, we need to find the values #y# can take in #y=x^2-1#.

One way to do this is to solve for #x# in terms of #y#: #x=+-sqrt(y+1)#.

Since #y+1# is under the square root sign, it must be the case that #y+1≥0#. Solving for #y# here, we get #y≥-1#.

In other words, the range is #{y|y≥-1}#.