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

1 Answer
Feb 11, 2018

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

Explanation:

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=2and 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:

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