homework4: problem 6\n(1 point)\nlet $f = \\sqrt{x + 3}$ and find the values below\n1. $f(x + h)=$\n2. $(f(x…

Explanation:

Step1: Substitute (x + h) into (f(x))

Replace (x) with (x + h) in (f(x)=\sqrt{x + 3}). So (f(x + h)=\sqrt{(x + h)+3}=\sqrt{x+h + 3})

Step2: Calculate (f(x + h)-f(x))

(f(x + h)-f(x)=\sqrt{x + h+3}-\sqrt{x + 3})

Step3: Find the limit

[ \begin{align*} \lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}&=\lim_{h\rightarrow0}\frac{\sqrt{x + h+3}-\sqrt{x + 3}}{h}\ &=\lim_{h\rightarrow0}\frac{(\sqrt{x + h+3}-\sqrt{x + 3})(\sqrt{x + h+3}+\sqrt{x + 3})}{h(\sqrt{x + h+3}+\sqrt{x + 3})}\ &=\lim_{h\rightarrow0}\frac{(x + h+3)-(x + 3)}{h(\sqrt{x + h+3}+\sqrt{x + 3})}\ &=\lim_{h\rightarrow0}\frac{h}{h(\sqrt{x + h+3}+\sqrt{x + 3})}\ &=\lim_{h\rightarrow0}\frac{1}{\sqrt{x + h+3}+\sqrt{x + 3}}\ &=\frac{1}{2\sqrt{x + 3}} \end{align*} ]

Step4: Find the equation of the tangent - line

First, when (x = 3), (f(3)=\sqrt{3 + 3}=\sqrt{6}) The slope of the tangent - line at (x = 3) is (m=\frac{1}{2\sqrt{3 + 3}}=\frac{1}{2\sqrt{6}}) Using the point - slope form (y - y_1=m(x - x_1)) with ((x_1,y_1)=(3,\sqrt{6})) (y-\sqrt{6}=\frac{1}{2\sqrt{6}}(x - 3)) (y=\frac{1}{2\sqrt{6}}x-\frac{3}{2\sqrt{6}}+\sqrt{6}) (y=\frac{1}{2\sqrt{6}}x+\frac{-3 + 12}{2\sqrt{6}}) (y=\frac{1}{2\sqrt{6}}x+\frac{9}{2\sqrt{6}}=\frac{1}{2\sqrt{6}}x+\frac{3\sqrt{6}}{4})

Answer:

1. (\sqrt{x + h+3})
2. (\sqrt{x + h+3}-\sqrt{x + 3})
3. (\frac{1}{2\sqrt{x + 3}})
4. (y=\frac{1}{2\sqrt{6}}x+\frac{3\sqrt{6}}{4})