### Instruction: Given the functions $f(x)=\dfrac{x+5}{3}$ and $g(x)=\dfrac{1}{f^{-1}(x)+1}$, find the value of $g(3)$.### Response: To find the value of $g(3)$, I need to first find the inverse of $f$, which is $f^{-1}$.
To find $f^{-1}$, I can switch $x$ and $y$ in the equation for $f$ and solve for $y$.
So, $y=\dfrac{x+5}{3}$ becomes $x=\dfrac{y+5}{3}$.
Multiplying both sides by $3$, I get $3x=y+5$.
Subtracting $5$ from both sides, I get $y=3x-5$.
This is the inverse of $f$, so $f^{-1}(x)=3x-5$.
Now, I can plug this into the equation for $g$ and simplify.
So, $g(x)=\dfrac{1}{f^{-1}(x)+1}$ becomes $g(x)=\dfrac{1}{3x-5+1}$.
Combining like terms in the denominator, I get $g(x)=\dfrac{1}{3x-4}$.
Finally, I can plug in $x=3$ and evaluate.
So, $g(3)=\dfrac{1}{3(3)-4}$.
Simplifying, I get $g(3)=\dfrac{1}{9-4}$.
This is the same as $g(3)=\dfrac{1}{5}$.