### Instruction: Suppose a function $f(x)$ has domain $(-\infty,\infty)$ and range $[-11,3]$. If we define a new function $g(x)$ by $$g(x) = f(6x)+1,$$ then what is the range of $g(x)$? Express your answer in interval notation.### Response: To find the range of $g(x)$, I need to see how the transformations affect the range of $f(x)$.
First, I notice that $g(x)$ is obtained by replacing $x$ with $6x$ inside $f$.
This means that $g(x)$ is horizontally compressed by a factor of 6 compared to $f(x)$.
That is, $g(x)$ will reach the same outputs as $f(x)$, but for smaller inputs.
However, this horizontal compression does not affect the range of $g(x)$.
It only affects the domain.
The range of $g(x)$ is still determined by the lowest and highest outputs that $f(x)$ can produce, which are -11 and 3, respectively.
Next, I see that $g(x)$ is obtained by adding 1 to the output of $f(x)$.
This means that $g(x)$ is vertically shifted up by 1 unit compared to $f(x)$.
That is, $g(x)$ will be 1 unit higher than $f(x)$ for any input.
This vertical shift does affect the range of $g(x)$.
It shifts the range of $f(x)$ up by 1 unit as well.
That means the lowest output of $g(x)$ will be -11 + 1 = -10, and the highest output of $g(x)$ will be 3 + 1 = 4.
Therefore, the range of $g(x)$ is the interval [-10, 4].