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To ensure that the function \( f(x) \) is continuous at \( x = 5 \), the left-hand limit and the right-hand limit of \( f(x) \) at \( x = 5 \) must be equal to the function's value at \( x = 5 \). Here's how we approach this problem step-by-step:

### Step 1: Left-hand limit
For \( x < 5 \), we are given:
\[
f(x) = b - 2x
\]
The left-hand limit of \( f(x) \) as \( x \) approaches 5 is:
\[
\lim_{x \to 5^-} f(x) = b - 2(5) = b - 10
\]

### Step 2: Right-hand limit
For \( x \geq 5 \), we are given:
\[
f(x) = \frac{150}{x - b}
\]
The right-hand limit of \( f(x) \) as \( x \) approaches 5 is:
\[
\lim_{x \to 5^+} f(x) = \frac{150}{5 - b}
\]

### Step 3: Continuity at \( x = 5 \)
For \( f(x) \) to be continuous at \( x = 5 \), we need the left-hand limit to equal the right-hand limit:
\[
b - 10 = \frac{150}{5 - b}
\]

### Step 4: Solve for \( b \)
Multiply both sides of the equation by \( 5 - b \) to eliminate the denominator:
\[
(b - 10)(5 - b) = 150
\]
Expand the left-hand side:
\[
(b - 10)(5 - b) = 5b - b^2 - 50 + 10b = -b^2 + 15b - 50
\]
Thus, the equation becomes:
\[
-b^2 + 15b - 50 = 150
\]
Move all terms to one side:
\[
-b^2 + 15b - 200 = 0
\]
Multiply through by -1 to simplify:
\[
b^2 - 15b + 200 = 0
\]
Now, solve this quadratic equation using the quadratic formula:
\[
b = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(1)(200)}}{2(1)}
\]
\[
b = \frac{15 \pm \sqrt{225 - 800}}{2}
\]
\[
b = \frac{15 \pm \sqrt{-575}}{2}
\]
Since the discriminant is negative, this equation has no real solutions, meaning that the approach we have taken to solve the problem does not lead to real solutions for \(b\).

Would you like to try understanding this approach

Find the two values of b for which f is a continuous function at 5.

Learn how to find the values of b that make a piecewise function continuous at x = 5. This problem involves limits and solving a quadratic equation, suitable for students in Grades 11-12.

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