Math Problem Statement

For what values of m and b is J(x) defined as continuous at x = 3 and x = 5?

Solution

We are tasked with finding the values of mm and bb such that the given piecewise function J(x)J(x) is continuous at x=3x = 3 and x=5x = 5.

The function is defined as:

\begin{cases} -2 & \text{if } x \leq 3 \\ mx + b & \text{if } 3 < x < 5 \\ 4 & \text{if } x \geq 5 \end{cases}$$ ### Continuity at $$x = 3$$: For the function to be continuous at $$x = 3$$, the left-hand limit (as $$x \to 3^-$$) and the right-hand limit (as $$x \to 3^+$$) must be equal to $$J(3)$$. 1. Left-hand limit: $$\lim_{x \to 3^-} J(x) = -2$$ 2. Right-hand limit (using the middle piece $$mx + b$$): $$\lim_{x \to 3^+} J(x) = m(3) + b = 3m + b$$ For continuity at $$x = 3$$, we need: $$-2 = 3m + b$$ ### Continuity at $$x = 5$$: For the function to be continuous at $$x = 5$$, the left-hand limit (as $$x \to 5^-$$) and the right-hand limit (as $$x \to 5^+$$) must be equal to $$J(5)$$. 1. Left-hand limit (using the middle piece $$mx + b$$): $$\lim_{x \to 5^-} J(x) = m(5) + b = 5m + b$$ 2. Right-hand limit: $$\lim_{x \to 5^+} J(x) = 4$$ For continuity at $$x = 5$$, we need: $$5m + b = 4$$ ### Solving the system of equations: We now have the system of two equations: 1. $$3m + b = -2$$ 2. $$5m + b = 4$$ Subtract equation 1 from equation 2: $$(5m + b) - (3m + b) = 4 - (-2 a)$$ $$2m = 6$$ $$m = 3$$ Substitute $$m = 3$$ into $$3m + b = -2$$: $$3(3) + b = -2$$ $$9 + b = -2$$ $$b = -11$$ ### Final answer: The values of $$m$$ and $$b$$ that make the function continuous at $$x = 3$$ and $$x = 5$$ are: $$m = 3, \quad b = -11$$ Would you like more details or explanations about the solution? Here are some related questions to explore further: 1. What are the conditions for a function to be continuous at a point? 2. How do you check differentiability of a piecewise function? 3. Can you find the derivative of the given function for $$3 < x < 5$$? 4. What happens if a piecewise function is not continuous at a given point? 5. Can a function be continuous but not differentiable at a point? **Tip:** Continuity at a point requires that the left-hand limit, right-hand limit, and function value all match at that point.

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Math Problem Analysis

Mathematical Concepts

Piecewise Functions
Continuity
Limits

Formulas

Left-hand limit: lim(x -> 3^-) J(x) = -2
Right-hand limit: lim(x -> 3^+) J(x) = m(3) + b
System of equations: 3m + b = -2, 5m + b = 4

Theorems

Continuity Theorem
Limit Theorem

Suitable Grade Level

Grade 11-12 or Early College Level