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Let \(f(x) = \begin{cases} 3x^2 + 2&\text{if } x\le 3, \\ ax - 1 &\text{if } x>3. \end{cases} \)  find \(a\)  if the graph of \(y=f(x)\) is continuous (which means the graph can be drawn without lifting your pencil from the paper).

 May 25, 2020
 #1
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The answer is a = 5.

 May 25, 2020
 #2
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I am not sure how Guest arrived at a=5, but that answer is incorrect.

 

In order for a function to be continuous at x=c, the function must fulfill three criteria. They are the following:

     1) \(f(c)\) is defined.

     2) \(\lim_{x\rightarrow c}f(x)\) is defined.

     3) \(\lim_{x\rightarrow c}f(x)=f(c)\)

 

Right now, only one criterion is met currently. \(f(3)\) is defined in this piecewise function because \(x=3\) is included in the domain. Let's move on to the next condition.

 

In order for \(\lim_{x\rightarrow c}f(x)\) to be defined, we have to ensure that left-side and right-side limit approach the same value. We set both limits equal to each other and find possible values for a, if one exists.

\(\lim_{x\rightarrow 3^-}f(x)=\lim_{x\rightarrow 3^+}f(x)\\ 3*3^2+2=3a-1\\ 29=3a-1\\ 3a=30\\ a=10 \)

 

At \(a=10\), both limits approach the same value. Therefore, we have met the second condition.

 

Finally, we have to make sure that the limit and the true value are not incongruous.

 

\(\lim_{x\rightarrow 3}f(x)\stackrel{?}{=}f(3)\\ 29=29\)

 

Yes, that means that a=10 makes the function continuous.


 

 May 25, 2020

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