Expert: Alon Mandes - 4/28/2009

Question
hi sir... can you help me from this problem... i cant solve this during my exam on calculus yesterday.... i think you can do this ... thanks very much here are some of the problems...

1. let f be a function define as

f(x) = 3x + 6a   if x < -3
    = 3ax - 7b  if -3 ≤ x ≤ 3
    = x - 12b   if x > 3
this is a piecewice function

Find the value of a and b that will make f continuous everywhere.

2. evaluate the limit.

([x]-1)/([x]-x) as x -> 1 raise to negative.... id ont understand why 1 raise to negative... pls explain to me...thanks a lot

Answer
Continuousity means :
Lim   f(x)  =  Lim  f(x)   =  Lim f(x)
x->α+          x=α-           x=α
Where x->α- is called : Lim when x reaches α from the left .. It's
the same as " x->α raise to negative " .
So,
As for our function f(x):
Lim f  =  3-12b
x->3+
Lim f  =  9a-7b
x->3-
Now, we want both limits to be equal :
3-12b=9a-7b -> 3=9a+5b : Eq(1)
& :
Lim f  =  9a-7b
x->-3+
Lim f  =  9+6a
x->-3-
Now, we want both limits to be equal :
9+6a=9a-7b -> 9=3a-7b : Eq(2)
All we have to do now, is solve a & b from both equations (1)&(2).

----------------------------------
Lim      ([x]-1)/([x]-x)  = (0-1)/(0-1)=1 .
x->1-

Alon.