The function is $$h(x)=\begin{cases} \dfrac{4x^2+5x-6}{x+2} & \text{if $x\neq-2$}, \\[6pt] 3x+k & \text{if $x= -2$}. \\ \end{cases}$$

After factoring the fraction I am left with $4x-3= -11$ when $x=-2$. If I multiply $(x+2)(4x-3)$ and then plug $-2$ into the resulting equation I am left with $-24$. If I do $4(-2)-3=3(-2)+k$ I am left with $k=-5$ which is none of the answer choices.

The answer choices to this question are

A. $k=20$

B. $k=0$

C. $k=2$

D. $k=4$

I am not sure where I go from here so any helpful tips on where I should start next would be appreciated.

  • $\begingroup$ Your reasoning is correct, and $k=-5$ is the right answer. $\endgroup$ – Olivier Oloa Jun 5 '16 at 21:04
  • $\begingroup$ Your reasoning is correct (except I'm not sure why you evaluated $(x+2)(4x-3)$ at $x=-2$, or how you got $-24$). The answer should be $k=-5$. $\endgroup$ – kccu Jun 5 '16 at 21:06
  • $\begingroup$ So do you just think that there was an error when writing the answer choices by the instructor? $\endgroup$ – user344249 Jun 5 '16 at 21:06

The right value is $k=-5$ as you have found.

Observing that, as $x \neq-2$, you have $$ f(x)=\frac{4x^2+5x-6}{x+2}=\frac{(4x-3)(x+2)}{x+2}=4x-3 $$ which is equal to $-11$ as $x \to -2$. On the other hand putting $x=-2$ in $3x+k$ gives $-6+k$ then you will need that $$ -6+k=-11 $$ that is $k=-5$.


Using the rule of l'Hospital on $\lim_{x\to -2}\frac{4x^2+5x-6}{x+2}$ we obtain $\lim_{x\to -2}8x+5=-11$. For none of the given options it is possible that $f$ is continuous since $-6+k \neq -11$ for $k\in\{0,2,4,20\}$.

  • $\begingroup$ So if none of them are possible it cannot be 0 either correct? This may be a dumb question but I do not want to overthink this $\endgroup$ – user344249 Jun 5 '16 at 21:15
  • $\begingroup$ For $k=0$ and $x=-2$ we obtain $-6$ which is clearly not $-11$. $\endgroup$ – Tesla Jun 5 '16 at 21:17

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