Let f(x) = x^2 + 2x - 3 for x < 2 and f(x) = ax + 1 for x >= 2. Determine the value of a that makes f continuous at x = 2.

Step-by-step solution

1. 1

   Write down what continuity at a point requires. A function is continuous at $x=2$ exactly when all three of these exist and agree:

   $\lim_{x\to 2^-}f(x)=f(2)=\lim_{x\to 2^+}f(x)$

   Each piece is a polynomial, so it is continuous on its own side; the only thing that can go wrong is a mismatch at the seam $x=2$.

2. 2

   Compute the left-hand limit from the first piece. For $x<2$ the rule is $f(x)=x^2+2x-3$, a polynomial, so the limit is just its value at $2$:

   $\lim_{x\to 2^-}f(x)=2^2+2(2)-3=4+4-3=5$

3. 3

   Compute the value and the right-hand limit from the second piece. The second rule applies for $x\ge 2$, so it supplies both $f(2)$ and the right-hand limit:

   $f(2)=2a+1,\qquad \lim_{x\to 2^+}f(x)=2a+1$

   Because the inequality is $x\ge 2$ rather than $x>2$, the function is already defined at the seam — there is no hole to fill, only a jump to close.

4. 4

   Set the two sides equal and solve. Continuity forces

   $2a+1=5\qquad\Longrightarrow\qquad 2a=4\qquad\Longrightarrow\qquad \boxed{a=2}$

5. 5

   Verify by rewriting the function. With $a=2$,

   $f(x)=\begin{cases} x^2+2x-3, & x<2\\ 2x+1, & x\ge 2\end{cases}$

   Approaching from the left, $f(1.99)=4.9401$; from the right, $f(2.01)=5.02$; and $f(2)=5$. The graph joins with no jump. Note that continuity is all that is achieved: the left derivative is $2x+2=6$ at $x=2$ while the right derivative is $2$, so the graph still has a corner there — no value of $a$ can make it differentiable.