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Suppose that a and b are positive real numbers, and let \( f(x) = \begin{cases} \frac{a}{b}x & \text{ if }x\le-4, \\ abx^2 & \text{ if }x>-4. \end{cases} \)if \(f(-4)=-\frac{60}{13} and f(4)=3120\), what is a+b?

 Jul 13, 2020
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Since  f(-4)  =  -60/13 

and     f(-4)  =  (a/b)x,

then a possibility is:   f(-4)  = (15/13)(-4)   --->   a = 15  and  b = 13

[other possibilities are multiples of  (15/13)]

 

Let's see what happens with f(4)   --->   f(4)  =  a·b·(4)2  =  (15)·(13)·42  =  3120.

 

Since this checks out,  a = 15  and  b = 13.

 Jul 13, 2020

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