MHB Find the values of a and b so the function is continuous everywhere.

[rayne1]

Find the values of a and b that make f continuous everywhere.

f(x) = (x2 − 4)/(x − 2)...if x < 2
...ax2 − bx + 3... if 2 ≤ x < 3
...4x − a +b....if x ≥ 3

This is a piece-wise function.

So I know that to be continuous everywhere, the function has to be one solid line. But I have no idea how to find a and b.

 

[johng1]

You must apply the definition of continuity. The limit of your function as x approaches 2 from the left is 4, the limit of the function as x approaches 2 from the right is $4a-2b+3$. So to be continuous at 2, it must be that $4=4a-2b+3$. Now do a similar calculation at 3 to get $9a-3b+3=12-a+b$. Solve for a and b.

 

[Evgeny.Makarov]

You can experiment with sliders in the following graph (click it).

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Then equate the values of the functions at the ends of their domains. Since $\frac{x^2-4}{x-2}$ is not define at $x=2$, instead of its value you need to take $\lim_{x\to2}\frac{x^2-4}{x-2}$.

Note: In plain text, it's customary to write x^2 for $x^2$.