- #1

karush

Gold Member

MHB

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$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$

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)

$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$

$x-2$ will cancel out

then if $x=2$ for $2(2)+1 = 5$ which is (E)

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