$$ f(x)=\begin{cases} \frac{2x^2-3x-2}{2x+1} & \text{for } x \neq -\frac{1}{2} \\ \small k & \text{for } x = -\frac{1}{2} \end{cases} $$

Let \(f\) be the function defined above. For what value of \(k\) is \(f\) continuous at \(x=-\dfrac{1}{2}\) ?

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The function \(f(x)=\dfrac{2x^2-3x-2}{2x+1}\) is undefined for \(x=-\dfrac{1}{2}\). It has a point discontinuity, but the limit can be found through factoring and cancelling out terms:

$$ f(x)=\frac{2x^2-3x-2}{2x+1} $$ $$ \lim_{x\to -\frac{1}{2}} \frac{2x^2-3x-2}{2x+1} $$ $$ =\lim_{x\to -\frac{1}{2}} \frac{(2x+1)(x-2)}{2x+1} $$ $$ =\lim_{x\to -\frac{1}{2}} x-2 $$ $$ = -\frac{1}{2}-2 $$ $$ = \boxed{-\frac{5}{2}} $$