Let $f(x)=\frac{3x^2+5x+8}{x^2-x+4}$ and $g(x)=x-1$. Find $f(g(x))+g(f(x))$, evaluated when $x=1$.
To evaluate $f(g(x))+g(f(x))$ when $x=1$, I need to plug in $x=1$ into both $f(g(x))$ and $g(f(x))$ and add them together. Let me start with $f(g(x))$. If $x=1$, then $g(x)=1-1=0$. So $f(g(x))$ becomes $f(0)$. To find $f(0)$, I need to plug in $x=0$ into the formula for $f(x)$. That gives me $f(0)=\frac{3(0)^2+5(0)+8}{(0)^2-(0)+4}=\frac{8}{4}=2$. So $f(g(x))=2$ when $x=1$. Now let me look at $g(f(x))$. If $x=1$, then $f(x)=\frac{3(1)^2+5(1)+8}{(1)^2-(1)+4}=\frac{16}{4}=4$. So $g(f(x))$ becomes $g(4)$. To find $g(4)$, I need to plug in $x=4$ into the formula for $g(x)$. That gives me $g(4)=4-1=3$. So $g(f(x))=3$ when $x=1$. Therefore, $f(g(x))+g(f(x))=2+3=5$ when $x=1$. 