Answer
$y=\displaystyle \frac{1}{64}x+\frac{11}{16}$
Work Step by Step
For x=$4$, the point on the graph is $(4, f(4))$.
The slope of the tangent at x=$4$ is $m=f^{\prime}(4).$
The point-slope equation of the tangent line is
$y-y_{1}=m(x-x_{1})$
$y-f(4)=f^{\prime}(4)(x-4)$
$f(4)=\displaystyle \frac{2+1}{2+2}=\frac{3}{4}$
For $f^{\prime}(x)$, use the quotient rule,
$\displaystyle \frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
$= \displaystyle \frac{(\sqrt{x}+1)^{\prime}(\sqrt{x}+2)-(\sqrt{x}+1)(\sqrt{x}+2)^{\prime}}{(\sqrt{x}+2)^{2}}$
$f^{\prime}(x)= \displaystyle \frac{(\frac{1}{2\sqrt{x}})(\sqrt{x}+2)-(\sqrt{x}+1)(\frac{1}{2\sqrt{x}})}{(\sqrt{x}+2)^{2}}$
... evaluate at x=$4$ ...
$f^{\prime}(4)=\displaystyle \frac{\frac{1}{4}(4)-(3)(\frac{1}{4})}{16}=\frac{4-3}{4\cdot 16}=\frac{1}{64}$
An equation for the tangent line is
$y-\displaystyle \frac{3}{4}=\frac{1}{64}(x-4)$
$y=\displaystyle \frac{1}{64}x-\frac{1}{16}+\frac{3}{4}$
$y=\displaystyle \frac{1}{64}x+\frac{11}{16}$
