SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given details
![\begin{gathered} w=g(z)=1+√(6-z) \\ (z,w)=(5,2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/mt575iireulfsg84ot1orfz3gdp9jlzqfl.png)
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:
![\begin{gathered} y=mx+b \\ where\text{ }m=f^(\prime)(a)\text{ }and\text{ }y|_(x=a)=f(a) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/e3r737r9a3f77s65m128qfj7xkxs3hdcor.png)
STEP 3: Find the tangent
We can first determine the slope of the tangent line by differentiating the function. Writing the function as:
![g(z)=1+(6-z)^{(1)/(2)}](https://img.qammunity.org/2023/formulas/mathematics/college/dyxrk9om0y3xpsra9ky59efcela6sbnjsj.png)
We find the derivative using the chain rule:
![\begin{gathered} \mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g' \\ =(d)/(dz)\left(1\right)+(d)/(dz)\left(\left(6-z\right)^{(1)/(2)}\right) \\ (d)/(dz)\left(1\right)=0,(d)/(dz)\left(\left(6-z\right)^{(1)/(2)}\right)=-\frac{1}{2\left(6-z\right)^{(1)/(2)}} \\ \\ =0-\frac{1}{2\left(6-z\right)^{(1)/(2)}} \\ =-\frac{1}{2\left(-z+6\right)^{(1)/(2)}} \\ \\ =-(1)/(2)(6-z)^{-(1)/(2)} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/xjhcfixwqgct3iqfbcr3nj2y1hzvclgh5b.png)
At the point of tangency, the slope is
![\begin{gathered} g^(\prime)(5)=-(1)/(2)(6-5)^{-(1)/(2)} \\ g^(\prime)(5)=-(1)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/sl8pdfli7ekmpflxtqvh39bh9e06q4jj0m.png)
We can begin to write the equation of the tangent line in slope-intercept form
![w=-(1)/(2)z+b](https://img.qammunity.org/2023/formulas/mathematics/college/hjp0re8onk34riqlb243tcue9bmkhf4nmt.png)
Substituting both coordinates of the point of tangency allows us to solve for the intercept:
![\begin{gathered} 2=-(1)/(2)(5)+b \\ -(1)/(2)\left(5\right)+b=2 \\ -(1)/(2)\cdot \:5+b=2 \\ -(5)/(2)+b=2 \\ (5)/(2)+b+(5)/(2)=2+(5)/(2) \\ b=(9)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/eiyqjy56isvb9c2hxhtg9hi81wct4eys6k.png)
The complete equation of the tangent line at (5,2) is then
![w=-(1)/(2)z+((9)/(2))](https://img.qammunity.org/2023/formulas/mathematics/college/j7bm60in6cmd1lnr0fo0qugyoalal8y5jq.png)