Question

homework4: problem 2 (1 point) let $f = sqrt{2x + 3}$ and find the values below 1. $f(x + h)=$ 2. $(f(x + h)-f(x))=$ 3. $lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=$ 4. find the equation of the line tangent to the graph of $f$ at $x = 2$. $y=$

Explanation:

Step1: Find $f(x + h)$

Substitute $x+h$ into $f(x)$.
$f(x + h)=\sqrt{2(x + h)+3}=\sqrt{2x+2h + 3}$

Step2: Find $f(x + h)-f(x)$

$f(x + h)-f(x)=\sqrt{2x+2h + 3}-\sqrt{2x + 3}$

Step3: Find $\lim_{h

ightarrow0}\frac{f(x + h)-f(x)}{h}$
Rationalize the numerator:
\[

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

\]
Take the limit as $h
ightarrow0$: $\lim_{h
ightarrow0}\frac{2}{\sqrt{2x+2h + 3}+\sqrt{2x + 3}}=\frac{2}{2\sqrt{2x + 3}}=\frac{1}{\sqrt{2x + 3}}$

Step4: Find the equation of the tangent - line at $x = 2$

First, find $f(2)=\sqrt{2\times2+3}=\sqrt{7}$
The slope of the tangent - line at $x = 2$ is $m=\frac{1}{\sqrt{2\times2+3}}=\frac{1}{\sqrt{7}}$
Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(2,\sqrt{7})$ and $m=\frac{1}{\sqrt{7}}$
$y-\sqrt{7}=\frac{1}{\sqrt{7}}(x - 2)$
$y=\frac{1}{\sqrt{7}}x-\frac{2}{\sqrt{7}}+\sqrt{7}=\frac{1}{\sqrt{7}}x+\frac{- 2 + 7}{\sqrt{7}}=\frac{1}{\sqrt{7}}x+\frac{5}{\sqrt{7}}$

Answer:

1. $\sqrt{2x+2h + 3}$
2. $\sqrt{2x+2h + 3}-\sqrt{2x + 3}$
3. $\frac{1}{\sqrt{2x + 3}}$
4. $y=\frac{1}{\sqrt{7}}x+\frac{5}{\sqrt{7}}$