Question

Ive been trying to solve this problem for forever!!! Pls help

Answer 1

For the function to be continuous at any x-value you need the left-hand limit to match the right-hand limit to match the function's value at that x-value.

For example, for the function to be continuous at x=2:

`\lim_(x \to 2^-) (x^2-4)/(x-2)` must equal
`\lim_(x \to 2^-) \left(ax^2 - bx-16 \right)`

This must also equal
`f(2) = a(2)^2 -b(2)-16` or
`f(2) = 4a-2b-16`.

So start by finding the first limit that has no a's or b's in it and set that equal to 4a-2b-16.

The problem is that this is only one equation and there are two variables, so we need a second equation to set up to be able to solve for a and b.

So, you need to repeat that whole process with the pieces on either side of x=3. We need to have:

`\lim_(x \to 3^-) \left(ax^2-bx-16\right) = \lim_(x \to 3^+) \left(10x -a+b \right) = f(3)`

That will give you a second equation with a's and b's. Once you have that, you'll have a system which you can solve using substitution or elimination.