Question 1021099
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For any value of *[tex \Large x \ =\ \alpha] where *[tex \Large \alpha\ \not = 2], *[tex \Large \lim_{x\right\alpha}\ =\ f(\alpha)] regardless of the value of *[tex \Large a].  There are a couple of basic limit theorems that you can apply to the segments of this function where *[tex \Large x\ \not =\ 2], and I'll leave that part of the analysis to you.


Therefore, for this piecewise fuction to be continuous for all reals, it is sufficient to find a number *[tex \Large a] such that *[tex \Large ax^2\ =\ x\ +\ 4] when *[tex \Large x\ =\ 2].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ =\ 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ \frac{3}{2}]


And the desired function is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \matrix \left\{\begin{array}{ccc} \frac{3}{2}x^2 & if & x\ <\ 2 \\ x\ +\ 4 & if & x\ \geq\ 2 \end{array}]


 *[illustration continuous_piecewise_function.jpg]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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