What are the values for a & b that make f continuous? f(x)=#(x^2-4)/(x-2)# if #x<=2#; f(x)=#ax^2+bx+3# if 2<x<3; f(x)=2x-a+b if #x>=3#

We have:

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

The issue of continuity will be focused on the intersection points between the various function definitions. i.e. #x=2#and #x=3#.

#x=2#:

Using the appropriate definition of #f(x)#, when #x=2#

# f(x) = (x^2-4)/(x-2) #

We note that when #x=2# the function is undefined. It does however have a removable discontinuity at #x=2# and we can write:

# f(2) = lim_(x rarr 2) (x^2-4)/(x-2) #
# \ \ \ \ \ \ \ = lim_(x rarr 2) ((x+2)(x-2))/(x-2) #
# \ \ \ \ \ \ \ = lim_(x rarr 2) (x+2) #
# \ \ \ \ \ \ \ = 4 #

In order to ensure continuity at #x=2#, we require continuity between the two neighbouring function definitions, so that:

# lim_(x rarr 2^-) (x^2-4)/(x-2) = lim_(x rarr 2^+) ax^2+bx+3 #
# :. 4 = 4a+2b+3 #
# :. 4a+2b = 1 # ..... [A]

#x=3#:

Using the appropriate definition of #f(x)#, when #x=3#

# f(x) = 2x - a+ b #

And so we have:

# f(3) = 6-a+b #

In order to ensure continuity at #x=3#, we require continuity between the two neighbouring function definitions, so that:

# lim_(x rarr 3^-) ax^2+bx+3 = lim_(x rarr 3^+) 2x - a+ b #
# :. 9a+3b+3 = 6 - a+ b #
# :. 10a+2b = 3 # ..... [B]

We now have two equations, [A] and [B] in two unknowns #a# and #b#, and solving these equation simultaneously, we get:

# a = 1/3 \ \ \ #, and #b = -1/6#

And we can verify the solution graphically: