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The problem asks to find the two values of \( b \) for which the function \( f(x) \) is continuous at \( x = -1 \). The function \( f(x) \) is defined as:

\[
f(x) = 
\begin{cases} 
b - 2x & \text{if } x < -1 \\
\frac{6}{x - b} & \text{if } x \geq -1 
\end{cases}
\]

To ensure continuity at \( x = -1 \), the left-hand limit and right-hand limit at \( x = -1 \) must equal the function value at that point. Let's calculate these limits and set them equal to each other.

### 1. Left-hand limit (\( x \to -1^{-} \)):
For \( x < -1 \), the function is given by \( f(x) = b - 2x \). Therefore, the left-hand limit as \( x \to -1 \) is:

\[
\lim_{x \to -1^{-}} f(x) = b - 2(-1) = b + 2
\]

### 2. Right-hand limit (\( x \to -1^{+} \)):
For \( x \geq -1 \), the function is given by \( f(x) = \frac{6}{x - b} \). Therefore, the right-hand limit as \( x \to -1 \) is:

\[
\lim_{x \to -1^{+}} f(x) = \frac{6}{-1 - b} = \frac{6}{-1 - b}
\]

### 3. Continuity condition:
For \( f(x) \) to be continuous at \( x = -1 \), the left-hand limit must equal the right-hand limit. Thus, we set:

\[
b + 2 = \frac{6}{-1 - b}
\]

### 4. Solving for \( b \):
Now, let's solve this equation for \( b \).

Multiply both sides by \( -1 - b \) to get rid of the denominator:

\[
(b + 2)(-1 - b) = 6
\]

Expand both sides:

\[
-(b + 2)(b + 1) = 6
\]

Expanding the left-hand side:

\[
-(b^2 + 3b + 2) = 6
\]

Simplifying:

\[
-b^2 - 3b - 2 = 6
\]

Move everything to one side:

\[
b^2 + 3b + 8 = 0
\]

### 5. Solving the quadratic equation:
To solve \( b^2 + 3b + 8 = 0 \), we use the quadratic formula:

\[
b = \frac{-3 \pm \sqrt{3^2 - 4(1)(8)}}{2(1)}
\]

Simplifying:

\[
b = \frac{-3 \pm \sqrt{9 - 32}}{2}
\]
\[
b = \frac{-3 \pm \sqrt{-23}}{2}
\]

Since the discriminant is negative, the solutions are complex:

\[
b = \frac{-3 \pm i\sqrt{23}}{2}
\]

Thus, the two values of \( b \) are:

\[
b = \frac{-3 + i\sqrt{23}}{2} \quad \text{and} \quad b = \frac{-3 - i\sqrt{23}}{2}
\]

Among these two values, the one with the greater absolute value is typically the one with the positive imaginary part, i.e., \( b = \frac{-3 + i\sqrt{23}}{2} \).

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Do you want more details, or have any further questions?

Here are five related questions to extend your learning:
1. How do you compute the modulus (absolute value) of a complex number?
2. What does it mean for a function to be continuous at a point?
3. Can you solve this quadratic equation by factoring?
4. What happens if a function is discontinuous at a point? How does it affect graph behavior?
5. How can we generalize continuity conditions for piecewise functions?

**Tip:** When solving quadratic equations, always check the discriminant (\( b^2 - 4ac \)) to determine if the roots are real or complex.

Find the two values of b for which f(x) is a continuous function at -1, given f(x) = {b - 2x if x < -1, 6/(x - b) if x ≥ -1}.

Learn how to determine continuity for a piecewise function at x = -1 by solving a quadratic equation. This problem involves concepts like limits, continuity, and the quadratic formula. Suitable for advanced high school students or early calculus learners.

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