217 AP Calculus Exam continous function with k

[karush]

217

$f(x)=\begin{cases}
\dfrac{(2x+1)(x-2)}{x-2}, &\text{for } x\ne 2 \\[3pt]
k, &\text{for } x=2 \\
\end{cases}$$(A)\,0\quad(B)\,1\quad(C)\,2\quad(D)\,3\quad(E)\,5\quad$

Spoiler

ok not sure if this the standard way to do this but
$x-2$ will cancel out
then if $x=2$ for $2(2)+1 = 5$ which is (E)

 

[skeeter]

Re: 217 AP Caluclus Exam continuous function with k

A function $f(x)$ is continuous at $x=c$ if ...

1. $f(c)$ is defined.

2. \(\displaystyle \lim_{x \to c} f(x)\) exists.

3. \(\displaystyle \lim_{x \to c} f(x)= f(c)\)$f(2) = k$ is defined

\(\displaystyle \lim_{x \to 2} f(x) = 2x+1\) exists.

\(\displaystyle \lim_{x \to 2} f(x) = 2(2)+1 = 5 = k\)

 

[karush]

Thanks I'll add that to the MeWe post