Question

homework4: problem 6
(1 point)
let $f = \sqrt{x + 3}$ and find the values below

1. $f(x + h)=$
2. $(f(x + h)-f(x))=$
3. $\lim_{h\to0}\frac{f(x + h)-f(x)}{h}=$
4. find the equation of the line tangent to the graph of $f$ at $x = 3$.

$y=$

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

\[

$$\begin{align*} \lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}&=\lim_{h ightarrow0}\frac{\sqrt{x + h+3}-\sqrt{x + 3}}{h}\\ &=\lim_{h ightarrow0}\frac{(\sqrt{x + h+3}-\sqrt{x + 3})(\sqrt{x + h+3}+\sqrt{x + 3})}{h(\sqrt{x + h+3}+\sqrt{x + 3})}\\ &=\lim_{h ightarrow0}\frac{(x + h+3)-(x + 3)}{h(\sqrt{x + h+3}+\sqrt{x + 3})}\\ &=\lim_{h ightarrow0}\frac{h}{h(\sqrt{x + h+3}+\sqrt{x + 3})}\\ &=\lim_{h ightarrow0}\frac{1}{\sqrt{x + h+3}+\sqrt{x + 3}}\\ &=\frac{1}{2\sqrt{x + 3}} \end{align*}$$

\]

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