For real numbers $x$, let \[f(x) = \left\{
\begin{array}{cl}
x+2 &\text{ if }x>3, \\
2x+a &\text{ if }x\le 3.
\end{array}
\right.\]What must the value of $a$ be to make the piecewise function continuous (which means that its graph can be drawn without lifting your pencil from the paper)?
For the function to be continuous, both of the two expressions must have the same value when $x=3$. Therefore, $3+2=2(3)+a$. Solving, we get that $a=\boxed{-1}$.
-1
Level 5
Algebra
Problem: For real numbers $x$, let \[f(x) = \left\{
\begin{array}{cl}
x+2 &\text{ if }x>3, \\
2x+a &\text{ if }x\le 3.
\end{array}
\right.\]What must the value of $a$ be to make the piecewise function continuous (which means that its graph can be drawn without lifting your pencil from the paper)?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the value of \( a \) that ensures continuity of the function \( f(x) \), we need to analyze the behavior of the function at the point where the definition changes, which is at \( x = 3 \). 

Continuity at a point requires that the left-hand limit, the right-hand limit, and the function value at that point are all equal. 

1. Calculate the right-hand limit as \( x \) approaches 3 from the right (i.e., \( x > 3 \)):
   \[
   f(3^+) = 3 + 2 = 5.
   \]

2. Calculate the left-hand limit as \( x \) approaches 3 from the left (i.e., \( x \le 3 \)):
   \[
   f(3^-) = 2(3) + a = 6 + a.
   \]

3. The function value at \( x = 3 \) is given by the definition for \( x \le 3 \):
   \[
   f(3) = 6 + a.
   \]

4. For continuity at \( x = 3 \), we set the left-hand limit equal to the right-hand limit:
   \[
   6 + a = 5.
   \]

5. Solve for \( a \):
   \[
   a = 5 - 6 = -1.
   \]

Thus, the value of \( a \) that makes the function continuous at \( x = 3 \) is \( -1 \).<|END_OF_LATENT|> The final answer is $-1$.