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Tuesday, February 2, 2016

Extreme value 1

Question:
If the following equation holds,
\lim_{x \rightarrow 2}\frac{\sqrt{9+ax}+b}{x-2}=-2
Find a and b.

Answer:
Since the extreme value of the denominator becomes zero (\lim_{x\rightarrow 2}(x-2)=0), the numerator must also be zero.
\lim_{x \rightarrow 2}(\sqrt{9+ax}+b)=\sqrt{9+2a}+b=0    \\   \rightarrow  b = -\sqrt{9+2a}  
Then, plug b into the original equation and rationalize the numerator and denominator.
\begin{eqnarray*}  \lim_{x \rightarrow 2}\frac{\sqrt{9+ax}-\sqrt{9+2a}}{x-2}     &=& \lim_{x \rightarrow 2}\frac{\sqrt{9+ax}-\sqrt{9+2a}}{x-2}\times \frac{\sqrt{9+ax}+\sqrt{9+2a}}{\sqrt{9+ax}+\sqrt{9+2a}}  \\     &=& \lim_{x \rightarrow 2}\frac{(9+ax)-(9+2a)}{(x-2)(\sqrt{9+ax}+\sqrt{9+2a})} \\     &=& \lim_{x \rightarrow 2}\frac{a(x-2)}{(x-2)(\sqrt{9+ax}+\sqrt{9+2a})} \\     &=& \frac{a}{2\sqrt{9+2a}} \end{eqnarray*}
Now, this result is equal to -2. Thus,
\frac{a}{2\sqrt{9+2a}} = -2 \\    \rightarrow a = -4\sqrt{9+2a} \\    \rightarrow a^2 = 16(9+2a)  \\    \rightarrow a^2-32a-144 = (a-36)(a+4)=0  
The extreme value a = -4\sqrt{9+2a} obtained above indicates that a<0. Therefore, a=-4. We know b = -\sqrt{9+2a}, so b=-1.


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