find the value of the constant a that makes the function continuous

(x)={(a^(2)+4x,x<-1),(ax+16,x>=-1):}

To determine the value of the constant \( a \) that makes the function continuous, we require that the left-hand limit and the right-hand limit of the function at \( x = -1 \) are equal to the function's value at \( x = -1 \).

The given function is:

\[
f(x) = 
\begin{cases} 
a^2 + 4x & \text{if } x < -1, \\
ax + 16 & \text{if } x \geq -1.
\end{cases}
\]

The condition for continuity at \( x = -1 \) is:

\[
\lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x) = f(-1).
\]

### Step 1: Compute \( \lim_{x \to -1^-} f(x) \)
For \( x < -1 \), \( f(x) = a^2 + 4x \). So:
\[
\lim_{x \to -1^-} f(x) = a^2 + 4(-1) = a^2 - 4.
\]

### Step 2: Compute \( \lim_{x \to -1^+} f(x) \)
For \( x \geq -1 \), \( f(x) = ax + 16 \). So:
\[
\lim_{x \to -1^+} f(x) = a(-1) + 16 = -a + 16.
\]

### Step 3: Equate the limits
For the function to be continuous, the two limits must be equal:
\[
a^2 - 4 = -a + 16.
\]

### Step 4: Solve for \( a \)
Simplify the equation:
\[
a^2 + a - 20 = 0.
\]

Factorize:
\[
(a + 5)(a - 4) = 0.
\]

Thus:
\[
a = -5 \quad \text{or} \quad a = 4.
\]

### Step 5: Verify both solutions
Both values of \( a \) satisfy the condition for continuity. Therefore, the function is continuous for \( a = -5 \) or \( a = 4 \).

---

Would you like a detailed explanation of why both values of \( a \) work, or do you have further questions?

### Related Questions:
1. What is the definition of continuity at a point in a piecewise function?
2. How do you determine the limits for a piecewise function at a boundary point?
3. Can a piecewise function have more than one solution for continuity?
4. What happens if \( a \) is not -5 or 4 in this case?
5. Can the second derivative also be continuous at \( x = -1 \) for some values of \( a \)?

### Tip:
Always verify both the left-hand and right-hand limits to ensure a function is continuous at a boundary point in piecewise functions!

Find the value of the constant a that makes the function continuous: f(x) = {(a^2 + 4x, x < -1), (ax + 16, x >= -1)}

Learn how to find the value of the constant a that ensures the continuity of a piecewise function at x = -1. This problem involves limits, quadratic equations, and the continuity condition. Suitable for Grades 10-12 students.

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