I need some help with this question:
Find the values of $a$ and $b$ that make $f$ continuous everywhere.
$$f(x)=\begin{cases}
x^2 − 4/x-2, &\text{if }x < 2\\
ax^2-bx+1, &\text{if } 2 ≤ x ≤ 3\\
4x – a + b, &\text{if } x ≥ 3\end{cases}$$
I started by writing two expressions for $a$ and $b$ based upon the left and right limits of each piece's endpoints. But I ended up with $a=3/2$ and $b=2/3$. Where did I go wrong?
Best Answer
I've seen this problem before and it looks like this may be the function you are trying to make continuous here:
$f(x) = \begin{cases} \frac{x^2-4}{x-2} & \mbox{if } x<2 \\ ax^2-bx+1 & \mbox{if } 2\leq x<3 \\ 4x-a+b & \mbox{if } x\geq 3 \end{cases}$
To make sure $f$ is continuous at $x=2$ you want to use the definition of what it means to be continuous by solving $\lim_{x \to 2} f(x) = f(2).$ Since $f$ is different from the left and right side of $x=2$, you need to instead find each one sided limit.
$$\lim_{x \to 2^{-}} f(x) = f(2)$$
$$\lim_{x \to 2^{+}} f(x) = f(2).$$
First the left sided limit: $$\lim_{x \to 2^{-}} \frac{x^2-4}{x-2} = f(2)$$ $$\lim_{x \to 2^{-}} \frac{(x+2)(x-2)}{x-2} = f(2)$$ $$\lim_{x \to 2^{-}} (x+2) = f(2)$$ $$4=f(2)$$ $$4=a(2)^2-b(2)+1$$ $$4=4a-2b+1$$ $$3=4a-2b$$ If you do this with the right sided limit, you'll see that you end up with $4a-2b+1=4a-2b+1$, which doesn't really give you any useful information. Now you want to do the same thing to make sure $f$ is continuous at $x=3$ $$\lim_{x \to 3^{-}} f(x) = f(3)$$ $$\lim_{x \to 3^{-}} ax^2-bx+1 = 12-a+b$$ $$a(3)^2-b(3)+1 = 12-a+b$$ $$9a-3b+1 = 12-a+b$$ $$10a-4b=11$$
And in this case, the right sided limit won't contribute anything useful. Now you just need to solve the following system of equations:
$$3=4a-2b$$ $$10a-4b=11$$
Here's a link to a very detailed work-through of a very similar problem with a bit more explanation of why this works.
https://jakesmathlessons.com/limits/solution-find-the-values-of-a-and-b-that-make-f-continuous-everywhere/