Answer :
SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given details
[tex]\begin{gathered} w=g(z)=1+\sqrt{6-z} \\ (z,w)=(5,2) \end{gathered}[/tex]STEP 2: Explain Tangent line
Tangent Lines:
The tangent line to a function f(x) is a line that touches the graph of the function at one specific point of tangency where it has the same slope as the function at that point.
If x = a is the point of tangency, the tangent line can be written in slope-intercept form as:
[tex]\begin{gathered} y=mx+b \\ where\text{ }m=f^{\prime}(a)\text{ }and\text{ }y|_{x=a}=f(a) \end{gathered}[/tex]STEP 3: Find the tangent
We can first determine the slope of the tangent line by differentiating the function. Writing the function as:
[tex]g(z)=1+(6-z)^{\frac{1}{2}}[/tex]We find the derivative using the chain rule:
[tex]\begin{gathered} \mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g' \\ =\frac{d}{dz}\left(1\right)+\frac{d}{dz}\left(\left(6-z\right)^{\frac{1}{2}}\right) \\ \frac{d}{dz}\left(1\right)=0,\frac{d}{dz}\left(\left(6-z\right)^{\frac{1}{2}}\right)=-\frac{1}{2\left(6-z\right)^{\frac{1}{2}}} \\ \\ =0-\frac{1}{2\left(6-z\right)^{\frac{1}{2}}} \\ =-\frac{1}{2\left(-z+6\right)^{\frac{1}{2}}} \\ \\ =-\frac{1}{2}(6-z)^{-\frac{1}{2}} \end{gathered}[/tex]At the point of tangency, the slope is
[tex]\begin{gathered} g^{\prime}(5)=-\frac{1}{2}(6-5)^{-\frac{1}{2}} \\ g^{\prime}(5)=-\frac{1}{2} \end{gathered}[/tex]We can begin to write the equation of the tangent line in slope-intercept form
[tex]w=-\frac{1}{2}z+b[/tex]Substituting both coordinates of the point of tangency allows us to solve for the intercept:
[tex]\begin{gathered} 2=-\frac{1}{2}(5)+b \\ -\frac{1}{2}\left(5\right)+b=2 \\ -\frac{1}{2}\cdot \:5+b=2 \\ -\frac{5}{2}+b=2 \\ \frac{5}{2}+b+\frac{5}{2}=2+\frac{5}{2} \\ b=\frac{9}{2} \end{gathered}[/tex]The complete equation of the tangent line at (5,2) is then
[tex]w=-\frac{1}{2}z+(\frac{9}{2})[/tex]