Question

Please explain this I am really confused about this contunity problem.

Answer 1

In order for a function $f(x)$ to be continuous at a point $x=c$, it has to satisfy the criterion,

$\displaystyle\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=f(c)$

First, notice that $x$ means that $x\neq2$, so the "leftmost" piece can be simplified to

$\dfrac{x^2-4}{x-2}=\dfrac{(x-2)(x+2)}{x-2}=x+2$

Then we have

$f(x)=\begin{cases}x+2&\text{for }x$

Each of the pieces are polynomials, so they are continuous everywhere. This means we only need to worry about the endpoints of the pieces, so we need to find $a,b$ such that $f(x)$ is continuous at both $x=2$ and $x=3$.

• At $x=2$: by definition of $f$, we have $f(2)=4a-2b+3$. We also need

$\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2}(x+2)=4$

$\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2}(ax^2-bx+3)=4a-2b+3$

$\implies4a-2b+3=4$

• At $x=3$: by definition, $f(3)=12-a+b$, we need

$\displaystyle\lim_{x\to3^-}f(x)=\lim_{x\to3}(ax^2-bx+3)=9a-3b+3$

$\displaystyle\lim_{x\to3^+}f(x)=\lim_{x\to3}(4x-a+b)=12-a+b$

$\implies9a-3b+3=12-a+b$

Now solve for $a,b$:

$\begin{cases}4a-2b+3=4\\9a-3b+3=12-a+b\end{cases}\implies\begin{cases}4a-2b=1\\10a-4b=9\end{cases}\implies\boxed{a=\frac72,b=\frac{13}2}$