Math Problem Statement
For what value(s) of b does the function f have continuity, removable discontinuity, infinite discontinuity, or jump discontinuity at x = 1?
Solution
Let's analyze the piecewise function and the given questions based on the image:
The function is defined as:
\begin{cases} -x + b, & \text{if } x < 1 \\ 5, & \text{if } x = 1 \\ \frac{-20}{x - b} + 1, & \text{if } x > 1 \text{ (and } x \neq b\text{)} \end{cases}$$ ### (a) For what value(s) of $$b$$ is $$f$$ continuous at 1? Given that $$b = 6$$, this value ensures continuity at $$x = 1$$. This is correct since all parts of the function align when $$b = 6$$, satisfying the continuity condition at $$x = 1$$. --- ### (b) For what value(s) of $$b$$ does $$f$$ have a removable discontinuity at 1? A removable discontinuity happens when the limit exists but does not match the value of the function. To check for this: 1. **Left-hand limit**: $$\lim_{x \to 1^-} (-x + b) = -1 + b$$ 2. **Right-hand limit**: $$\lim_{x \to 1^+} \left( \frac{-20}{x - b} + 1 \right)$$ For a removable discontinuity, the limits must be equal but not necessarily equal to the function value $$f(1) = 5$$. We need to solve: $$-1 + b = 5 \quad \Rightarrow \quad b = 6$$ Thus, there is no removable discontinuity at $$x = 1$$ when $$b = 6$$, as this value makes the function continuous. So, **no other values of $$b$$ lead to a removable discontinuity**, as continuity is already ensured for $$b = 6$$. --- ### (c) For what value(s) of $$b$$ does $$f$$ have an infinite discontinuity at 1? An infinite discontinuity at $$x = 1$$ would occur if the function blows up to infinity as $$x \to 1^+$$. For that, we need the denominator $$x - b$$ to approach zero as $$x \to 1$$. This would occur if $$b = 1$$, since: $$\lim_{x \to 1^+} \frac{-20}{x - 1} = -\infty$$ So, the value of $$b$$ that creates an infinite discontinuity at $$x = 1$$ is $$b = 1$$. --- ### (d) For what value(s) of $$b$$ does $$f$$ have a jump discontinuity at 1? A jump discontinuity occurs if the left-hand and right-hand limits exist but are not equal. We already computed: - Left-hand limit: $$\lim_{x \to 1^-} (-1 + b) = b - 1$$ - Right-hand limit: $$\lim_{x \to 1^+} \frac{-20}{x - b} + 1$$ We solve for when these two limits are not equal and $$f(x)$$ jumps. To find the intervals for which the function has a jump discontinuity, we use a **process of elimination** based on the other cases (continuity, removable, and infinite discontinuities). Thus, **$$b \neq 6$$** and $$b \neq 1$$ could lead to a jump discontinuity. --- Would you like more details on any of the above solutions or need help with another question? ### Related Questions: 1. Why does $$b = 6$$ ensure continuity at $$x = 1$$? 2. What is the difference between removable and jump discontinuities? 3. How do limits help identify the type of discontinuity? 4. Could there be any other values of $$b$$ where the function is discontinuous? 5. How would you check if the function is differentiable at $$x = 1$$? ### Tip: Always check both the left-hand and right-hand limits carefully to classify discontinuities at a point.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Discontinuities (removable, infinite, jump)
Limits
Formulas
Limit from the left: lim(x -> 1-) f(x)
Limit from the right: lim(x -> 1+) f(x)
Function value: f(1)
Theorems
Continuity Definition
Discontinuity Classification
Suitable Grade Level
Grades 11-12 or College Calculus I
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