-apc.1.1.06

• MHB
Gold Member
MHB
If $f(x)=\begin{cases} \dfrac{(2x+1)(x-2)}{x-2}, &\text{for } x\ne 2 \\[3pt] k, &\text{for } x=2 \\ \end{cases}$
for what number k will the function be continuous
a. 0 b. 1 c. 2 d. 3 e. 5
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I chose
[e] x is undefined at $x=2$ so
rewrite as $2x+1$
plug in $f(2)=2(2)+1=5=k$

hopefully
probably suggestions etc

skeeter
for f(x) to be continuous at x = 2 ...

$\displaystyle \lim_{x \to 2} f(x) = f(2)$

Gold Member
MHB
So discontinuous. Means a hole always?

skeeter
So discontinuous. Means a hole always?

not just a hole (a removable discontinuity)

also, a vertical asymptote or a jump (non-removable discontinuities)