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: Key Approach:
For continuity at \(x=3\), the left-hand limit, right-hand limit and the function's value at \(x=3\) must all be equal.

Step 2: Computing the value and left-hand limit at x=3:
\(f(3)=a(3)+1=3a+1\); this is also the left-hand limit since the first branch applies for \(x\le3\).

Step 3: Computing the right-hand limit:
\(\displaystyle\lim_{x\to3^{+}}f(x)=b(3)+3=3b+3\).

Step 4: Equating and simplifying:
\(3a+1=3b+3\ \Rightarrow\ 3a-3b=2\ \Rightarrow\ a-b=\dfrac23\).

Final Answer:
Continuity at \(x=3\) requires exactly \(a-b=\dfrac23\), i.e. \(a=b+\dfrac23\) for any real \(b\) (one condition on two unknowns, giving a family of solutions).\[ \boxed{a-b=\dfrac23} \]