# Determine the values of a and b of the following function such that the function boundaries at x = -3 and x = 3 exist. {3x+6a, if x<-3f(x) = {3ax-7b, if -3≤x≤3 ...

Determine the values of a and b of the following function such that the function boundaries at x = -3 and x = 3 exist.

**{3x+6a, if x<-3****f(x) = {3ax-7b, if -3≤x≤3**** {x-12b, if x>3**

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`f(x) ={(3x+6a if x<-3),(3ax-7b if -3<=x<=3), (x-12b if x>3):}`

The boundary `x=-3` exists if the function at the interval `xlt-3` and the function at `-3lt=xlt=3` result to same value of f(x).

Also, the boundary x=3 exists if the function at `-3lt=xlt=3` and the function `xgt3` result to same value of f(x) too.

Base on these two conditions, values of a and b can now be solved.

To do so, set f(x) of `xlt-3` and f(x) of `-3lt=xlt=3` equal to each other.

`3x + 6a = 3ax - 7b`

And substitute the boundary between these intervals which is x=-3.

`3(-3) + 6a = 3a(-3)-7b`

`-9+6a=-9a-7b`

`-9=-15a-7b`

`9=15a+7b` (Let this be EQ1.)

Next, set f(x) of `-3lt=xlt=3 ` and f(x) of `xgt3` equal to each other.

`x-12b = 3ax-7b`

Substitute the boundary of these two intervasl which is x=3.

`3-12b=3a(3)-7b`

`3-12b=9a-7b`

`3=9a+5b` (Let this be EQ2.)

Then, use elimination method of system of equations. So, multiply EQ1 by 5 and EQ2 by -7, and add the resulting expressions to eliminate b.

`5(9=15a+7b)` ====> `45=75a+35b`

`(+) ` `-7(3=9a+5b)` ====> `(+)` `-21=-63a-35b`

`----------------`

`24=12a`

Then, solve for a.

`12a=24`

`a=24/12`

`a=2`

To solve for b, substitute value of a to either EQ1 and EQ2.

`a=2` , `3 = 9a+5b` `-15=5b`

`3=9*2+5b ` `-15/5=b`

`3=18+5b ` `-3=b`

**Hence, the boundaries x=-3 and x=3 exist if a=2 and b=-3 . **