Question

using the definition of derivative, find the derivative of $f(x)=\frac{1}{2x + 1}$.
$\frac{1}{(2x+1)^2}$
$-\frac{2}{(2x + 1)^2}$
$-(2x + 1)^{-2}$
$\frac{2}{(2x+1)}$
$\frac{1}{2}$
$-\frac{2}{(2x+1)}$
question 15 1 pts
which of the following is a correct statement about the continuity of $g(x)$
$g(x)=\begin{cases}\frac{(x + 3)(x - 1)}{(x + 3)}& \text{if }x
eq - 3\\5& \text{if }x=-3end{cases}$
$g(x)$ is discontinuous at $x=-3$ because although $lim_{x
ightarrow - 3}f(x)$ does not exist
$g(x)$ is continuous at $x=-3$
$g(x)$ is discontinuous at $x=-3$ because although $g(-3)$ and $lim_{x
ightarrow - 3}f(x)$ both exist, $lim_{x
ightarrow - 3}f(x)
eq g(-3)$

Explanation:

Step1: Recall derivative definition

The definition of the derivative of a function $y = f(x)$ is $f^\prime(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=\frac{1}{2x + 1}$, then $f(x+h)=\frac{1}{2(x + h)+1}=\frac{1}{2x+2h + 1}$.

Step2: Substitute into derivative formula

\[

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

\]

Step3: Evaluate the limit

As $h
ightarrow0$, we have $f^\prime(x)=-\frac{2}{(2x + 1)^2}=-2(2x + 1)^{-2}$.

for second - part:

Step1: Simplify the function for $x

eq - 3$
For $x
eq - 3$, $g(x)=\frac{(x + 3)(x - 1)}{(x + 3)}=x - 1$.

Step2: Find the limit as $x

ightarrow - 3$
$\lim_{x
ightarrow - 3}g(x)=\lim_{x
ightarrow - 3}(x - 1)=-3-1=-4$.

Step3: Compare with the function value at $x=-3$

$g(-3) = 5$. Since $\lim_{x
ightarrow - 3}g(x)=-4$ and $g(-3)=5$, $\lim_{x
ightarrow - 3}g(x)
eq g(-3)$.

Answer:

$-\frac{2}{(2x + 1)^2}$