Parameter that makes a piecewise function continuous
Given the piecewise function $$ f(x,\alpha) = \begin{cases} \alpha x^2 + x + 3, & \text{if } x \le 1, \\ \sin\left(\dfrac{\pi x}{2}\right) + 2\alpha, & \text{if } x > 1, \end{cases} $$ determine the value of $\alpha$ for which $f$ is continuous at $x=1$.
Explanation of Method
Because the first piece includes the boundary point ($x \le 1$), the value $f(1,\alpha)$ is determined by that formula, and continuity at $x=1$ reduces to a single condition: the right-hand limit (from the second piece) must equal $f(1,\alpha)$. Both pieces are built from continuous elementary functions (a polynomial and a sine), so each one-sided limit is found by straightforward substitution. Setting the two expressions equal produces a linear equation in $\alpha$.
- Evaluate $f(1,\alpha)$ using the first piece (valid at $x=1$ since $1\le1$).
- Evaluate the right-hand limit using the second piece.
- Set the two expressions equal and solve for $\alpha$.
Step 1 · Evaluate $f(1,\alpha)$
Since $x=1$ satisfies $x \le 1$, the function value there is given directly by the first piece:
$$ f(1,\alpha) = \alpha(1)^2 + 1 + 3 = \alpha + 4. $$
Because polynomials are continuous, this is also the left-hand limit:
$$ \lim_{x \to 1^-} f(x,\alpha) = \alpha + 4. $$
Step 2 · Evaluate the right-hand limit
For $x>1$, $f(x,\alpha) = \sin\left(\dfrac{\pi x}{2}\right) + 2\alpha$. Since $\sin$ is continuous, the right-hand limit at $x=1$ is found by substitution:
$$ \lim_{x \to 1^+} f(x,\alpha) = \sin\left(\frac{\pi}{2}\right) + 2\alpha = 1 + 2\alpha. $$
Step 3 · Set up and solve the continuity condition
For $f$ to be continuous at $x=1$, the right-hand limit must equal the value $f(1,\alpha)$ (which, since the first piece includes $x=1$, is automatically also the left-hand limit):
$$ \alpha + 4 = 1 + 2\alpha. $$
Solving for $\alpha$:
$$ 4 - 1 = 2\alpha - \alpha \quad\Longrightarrow\quad \alpha = 3. $$
Summary Table
| Quantity | Expression | Value at $\alpha=3$ |
|---|---|---|
| $f(1,\alpha)$ (and left limit) | $\alpha+4$ | $7$ |
| Right-hand limit | $1+2\alpha$ | $7$ |
Additional Notes
- Because the domain of the first piece already includes $x=1$ (note the $\le$), there is only one continuity condition to solve — the value from the first piece automatically serves as both $f(1,\alpha)$ and the left-hand limit, so they never need to be checked separately.
- Both pieces are continuous on their own domains (a polynomial and a sine function shifted by a constant), so the only possible discontinuity in $f$ can occur at the seam $x=1$ — this is why checking a single point resolves continuity for the whole function.
- At $\alpha=3$, both sides meet at the value $7$, giving a smooth-looking (though not necessarily differentiable) transition from the parabola to the sine curve at $x=1$.
Result
The function is continuous at $x=1$ when: