how can i find the domain of $f(x)=x^2+1$?

I don't understand how to find the domain ranges of this. I know that it is infinity but why? I know that it cannot equal zero but squaring a zero and adding one makes it 1 so that is a possible answer. The same goes for negatives. Squaring a negative and adding one makes it true.

3 Answers

I think you are confusing domains and ranges. The domain is the set the function takes values from, and the range is the set the function returns values from.

$f(x)=x^2+1$ is defined for all real numbers of course, so we would imagine that is the domain (to be pedantic, a function is not yet a function until a domain has been specified). To find its range, we consider the output the function can give.

If you draw a graph of $x^2+1$, you'll notice it is symmetric about the $y$ axis, and that its lowest point occurs at $x=0$, where we get $f(0)=1$, and that it grows unboundedly in both directions. This picture makes it 'obvious' that the range is then positive real numbers greater than or equal to 1, or $\{x\in\mathbb{R} : x\geq 1\}$

To put it in very simple (though not absolutely 100% accurate) terms, if you are looking at $y=f(x)$ the the domain means the set of all possible values for $x$, and the range means the set of all possible values for $y$.

The domain is simply those values for which this function is defined to output another value (and that output would be be in the functions range). So when you say that $0$ is in the domain since $f(0)=0^2+1=1$ but indeed $0$ isn't in the range since there is no real $x$ such that $x^2+1=0$.