Value of k that makes a function continuous at 2
Given the piecewise function $$ f(x) = \begin{cases} kx^2+1, & x<2, \\ 5, & x=2, \\ x+3, & x>2, \end{cases} $$ determine the value of $k$ (if any) for which $f$ is continuous at $x=2$.
Explanation of Method
Unlike a two-piece function where the boundary point's value is automatically inherited from one of the pieces, here $f(2)$ is separately defined to be $5$. Continuity at $x=2$ therefore requires three quantities to agree: the left-hand limit, the right-hand limit, and the explicitly assigned value $f(2)=5$. Since the right-hand piece is a simple linear function, its limit already equals $5$ automatically; the only remaining condition is to choose $k$ so that the left-hand piece also limits to $5$.
- Read off $f(2)$ directly from its explicit definition.
- Evaluate the right-hand limit and confirm it already matches $f(2)$.
- Evaluate the left-hand limit in terms of $k$, and set it equal to $f(2)$.
- Solve the resulting equation for $k$.
Step 1 · Read off $f(2)$
By the explicit middle case of the definition:
$$ f(2) = 5. $$
Step 2 · Evaluate the right-hand limit
For $x>2$, $f(x)=x+3$, a continuous polynomial, so:
$$ \lim_{x\to2^+} f(x) = \lim_{x\to2^+}(x+3) = 2+3 = 5. $$
This already matches $f(2)=5$, so the right-hand piece imposes no restriction on $k$.
Step 3 · Evaluate the left-hand limit in terms of $k$
For $x<2$, $f(x)=kx^2+1$, also a continuous polynomial (for any fixed $k$), so:
$$ \lim_{x\to2^-} f(x) = \lim_{x\to2^-}\left(kx^2+1\right) = k(2)^2+1 = 4k+1. $$
Step 4 · Set up and solve the continuity condition
For continuity at $x=2$, all three quantities must agree:
$$ \lim_{x\to2^-}f(x) = \lim_{x\to2^+}f(x) = f(2) \qquad\Longleftrightarrow\qquad 4k+1 = 5 = 5. $$
The right-hand equality ($5=5$) already holds automatically; solving the left-hand equality for $k$:
$$ 4k+1=5 \quad\Longrightarrow\quad 4k=4 \quad\Longrightarrow\quad k=1. $$
Summary Table
| Quantity | Expression | Value at $k=1$ |
|---|---|---|
| $f(2)$ | $5$ (given) | $5$ |
| $\displaystyle\lim_{x\to2^+}f(x)$ | $x+3$ | $5$ |
| $\displaystyle\lim_{x\to2^-}f(x)$ | $4k+1$ | $5$ |
Additional Notes
- Unlike problems where the boundary value is taken from one of the pieces (making continuity a single equality between two one-sided limits), here the explicitly assigned value $f(2)=5$ acts as a separate, third condition — though in this particular problem the right-hand piece happens to already satisfy it, leaving only one equation to solve for $k$.
- Had the right-hand limit not already equaled $5$ (for instance, if $f(x)=x+2$ for $x>2$, giving a limit of $4$), then no value of $k$ could make the function continuous at $x=2$, since the left-hand piece alone cannot be made to match both $f(2)=5$ and a mismatched right-hand limit simultaneously.
- This illustrates that continuity at a point where the function's value is prescribed independently (rather than inherited from either piece) generally imposes as many as three separate conditions, all of which happened to be simultaneously satisfiable here.
Result
The function is continuous at $x=2$ when: