Question

Find the values of \(a\) and \(b\) for which the function defined by \(f(x)=\begin{cases}ax+1, & x\le3\\bx+3,& x>3\end{cases}\) is continuous at \(x=3\).

Hint:

Match the left-hand and right-hand pieces at x=3 for continuity.

Answer 1

Step 1: Setting up the continuity condition directly:
Continuity means \(\displaystyle\lim_{x\to3^-}f(x)=\lim_{x\to3^+}f(x)=f(3)\).

Step 2: Plugging in x=3 into both branch formulas:
Branch 1 at \(x=3\): \(3a+1\). Branch 2's limit as \(x\to3^+\): \(3b+3\) (by continuity of the linear expression itself).

Step 3: Equating and solving for the relation between a and b:
\(3a+1=3b+3\Rightarrow a=b+\dfrac23\); so any pair \((a,b)\) satisfying this relation works, e.g. \(b=0,a=2/3\), or in general \(a-b=2/3\).

Final Answer:
\[ \boxed{a-b=\dfrac23} \]