When $f(x) = x^2+3x+1 (x\geq1), x+2(x<1)$, $g(t) = ㅣ3t-12ㅣ$, $t=f(x)$ In this case, how can $g(f(x))$ be define between when $f(x)= [3,5]$?
$f(x)$ do not have $3<f(x)\leq5$ as a result.
so $g(t)$ do not have $[3,5]$ for its domain. But when I plotted the graph of $g(f(x))$, it's even continuous for every real number. Can I know what I misunderstood?
The composition of $\ g\ $ and $\ f\ $ is continuous because $\ g\ $ is continuous everywhere, $\ f\ $ is continuous everywhere except at $\ x=1\ $, where it is continuous from above, and $\ \displaystyle g\left(\lim_{x\rightarrow1^-}f(x)\right)=$$g(3)=$$3=$$g(5)=$$g(f(1))\ $. Alternatively, by definition, \begin{align} g(f(x))&=|3f(x)-12|\\ &=\cases{|3x-6|&if $\ x<1$\\ \left|3x^2+9x-9\right|&if $\ x\ge1\ $,} \end{align} so the graph of $\ y=g(f(x))\ $ is that of $\ y= |3f(x)-12|\ $ for $\ x<1\ $, and it is that of $\ y= \left|3x^2+9x-9\right|\ $ for $\ x\ge1\ $. But when $\ x=1\ $ the graphs of $\ y= |3f(x)-12|\ $ and $\ y= \left|3x^2+9x-9\right|\ $ intersect (namely at the point $\ (1,3)\ $), so the graph of $\ y=g(f(x))\ $ will have no jumps.