### Instruction: Let $f(x)=3x-2$, and let $g(x)=f(f(f(f(x))))$.  If the domain of $g$ is $0\leq x\leq 2$, compute the range of $g$.### Response: To find the range of $g$, I need to figure out the minimum and maximum values that $g(x)$ can take when $x$ is between 0 and 2.
I could try to plug in some values of $x$ into $g$, but that would involve calculating $f$ four times for each $x$, which is tedious and might not give me the exact endpoints of the range.
A better idea is to try to simplify $g$ by finding a formula for $g(x)$ in terms of $x$, without using $f$.
To do that, I can use the definition of $f$ and substitute it into $g$.
So, $g(x)=f(f(f(f(x))))=f(f(f(3x-2)))$.
Then, I can apply $f$ again to the innermost expression, $3x-2$, and get $f(3x-2)=3(3x-2)-2=9x-8$.
So, $g(x)=f(f(f(3x-2)))=f(f(9x-8))$.
I can repeat this process until I get rid of all the $f$s.
Next, I get $f(9x-8)=3(9x-8)-2=27x-26$.
So, $g(x)=f(f(9x-8))=f(27x-26)$.
Finally, I get $f(27x-26)=3(27x-26)-2=81x-80$.
So, $g(x)=f(27x-26)=81x-80$.
Now, I have a simple linear function for $g(x)$, and I can use it to find the range.
The slope of $g$ is 81, which is positive, so $g$ is an increasing function.
That means that the minimum value of $g(x)$ occurs when $x$ is at the left endpoint of the domain, which is 0.
Plugging in $x=0$ into $g$, I get $g(0)=-80$.
Similarly, the maximum value of $g(x)$ occurs when $x$ is at the right endpoint of the domain, which is 2.
Plugging in $x=2$ into $g$, I get $g(2)=81(2)-80=82$.
Therefore, the range of $g$ is the interval from -80 to 82, or $[-80,82]$.