Question

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`

Answer 1

`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=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: