How do you find the values of 'a' and 'b' that make f(x) continuous everywhere given #(x^2 - 4)/(x - 2)# if x < 2, #ax^2 - bx + 3# if 2 < x < 3 and #2x - a + b# if #x>=3#?

1 Answer
Truong-Son N.
Jul 25, 2016

I'm guessing you have a piece-wise function:

#f(x) = {((x^2 - 4)/(x-2)",", x < 2),(ax^2 - bx + 3",", 2 < x < 3),(2x - a + b",", x >= 3):}#

The way the function is defined, we already know that #x in (-oo,2) uu (2,3) uu [3,oo)#. That is, #x ne 2# (but #x# can be #3#). However, because of that, there is no way for #f(x)# to be continuous everywhere.

#f(x)# would have a discontinuity (a hole) at #x = 2#, since even though you can cancel out #x - 2#, you still have #2 - 2 = 0#.

#:.# One might say that #f(x)# is continuous when #a,b# are real and finite, but only if #x ne 2#. So strictly speaking, #f(x)# cannot be continuous everywhere.

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