If $f(x)$ if a function whose domain is $(-\infty,\infty)$ and whose range is $[-11,3]$, and $g(x) = f(x)+1,$ then what is the range of $g(x)$? Express your answer in interval notation.
Since every output of $f(x)$ is increased by $1$, $g(x)$ obtains every value that $f(x)$ does plus $1$. Since the range of $f(x)$ is $[-11,3]$, the range of $g(x)$ is $\boxed{[-10,4]}$.
To find the range of $g(x)$, we need to consider the possible values that $f(x)$ can take in its range which is $[-11,3]$.
Since $g(x) = f(x) + 1$, every value in the range of $f(x)$ will be increased by $1$ to get the corresponding value in the range of $g(x)$.
Therefore, the range of $g(x)$ will be $[-11+1, 3+1]$, which simplifies to $[-10,4]$.
Hence, the range of $g(x)$ is $[-10,4]$ in interval notation.
To find the range of $g(x)$, we need to determine the possible values that $g(x)$ can take on.
We know that the range of $f(x)$ is $[-11,3]$, which means that $f(x)$ can take on any value between $-11$ and $3$, inclusive.
Now, since $g(x) = f(x) + 1$, we can conclude that $g(x)$ will have values that are $1$ unit greater than the corresponding values of $f(x)$. In other words, every value of $f(x)$ will be incremented by $1$ to get the corresponding value of $g(x)$.
So, the range of $g(x)$ will have each value from the range of $f(x)$ incremented by $1$.
In interval notation, this means that the range of $g(x)$ is $[-11+1, 3+1] = [-10, 4]$.