https://file.aiquickdraw.com/mathbot/file/17264084257629jsszsqv.jpg

Let's work through the problem step by step.

### The function is defined as:

\[
f(x) = 
\begin{cases} 
1 + mx & \text{if } x \leq 1, \\
4 - mx & \text{if } x > 1,
\end{cases}
\]
where \( m \) is a constant.

---

### Part (A): Graph \( f \) for \( m = 1 \), and find:

\[
\lim_{x \to 1^-} f(x) \quad \text{and} \quad \lim_{x \to 1^+} f(x)
\]

**Step 1: Evaluate \( \lim_{x \to 1^-} f(x) \)** (approaching from the left):

For \( x \leq 1 \), \( f(x) = 1 + mx \). With \( m = 1 \):
\[
f(x) = 1 + x.
\]
Thus, 
\[
\lim_{x \to 1^-} f(x) = 1 + 1 = 2.
\]

**Step 2: Evaluate \( \lim_{x \to 1^+} f(x) \)** (approaching from the right):

For \( x > 1 \), \( f(x) = 4 - mx \). With \( m = 1 \):
\[
f(x) = 4 - x.
\]
Thus, 
\[
\lim_{x \to 1^+} f(x) = 4 - 1 = 3.
\]

### Part (B): Graph \( f \) for \( m = 2 \), and find:

\[
\lim_{x \to 1^-} f(x) \quad \text{and} \quad \lim_{x \to 1^+} f(x)
\]

**Step 1: Evaluate \( \lim_{x \to 1^-} f(x) \)** (approaching from the left):

For \( x \leq 1 \), \( f(x) = 1 + mx \). With \( m = 2 \):
\[
f(x) = 1 + 2x.
\]
Thus, 
\[
\lim_{x \to 1^-} f(x) = 1 + 2(1) = 3.
\]

**Step 2: Evaluate \( \lim_{x \to 1^+} f(x) \)** (approaching from the right):

For \( x > 1 \), \( f(x) = 4 - mx \). With \( m = 2 \):
\[
f(x) = 4 - 2x.
\]
Thus, 
\[
\lim_{x \to 1^+} f(x) = 4 - 2(1) = 2.
\]

---

### Part (C): Find \( m \) so that \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \)

To make the left-hand and right-hand limits equal at \( x = 1 \), we set:
\[
1 + m(1) = 4 - m(1),
\]
which simplifies to:
\[
1 + m = 4 - m.
\]
Solving for \( m \):
\[
1 + m + m = 4,
\]
\[
2m = 3,
\]
\[
m = \frac{3}{2}.
\]

Thus, \( m = \frac{3}{2} \).

---

### Part (D): Verbal description of each graph

- For \( m = 1 \) (Part A), the function has a jump discontinuity at \( x = 1 \) because \( \lim_{x \to 1^-} f(x) = 2 \) and \( \lim_{x \to 1^+} f(x) = 3 \).
- For \( m = 2 \) (Part B), the function also has a jump discontinuity, but the limits are reversed: \( \lim_{x \to 1^-} f(x) = 3 \) and \( \lim_{x \to 1^+} f(x) = 2 \).
- For \( m = \frac{3}{2} \) (Part C), the function is continuous at \( x = 1 \), meaning \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = 2.5 \), so the graph is smooth without jumps at that point.

---

Let me know if you'd like any further clarification on these steps! Here are five related questions:

1. What is the general definition of a piecewise function?
2. How can you determine if a function is continuous at a point?
3. What are the conditions for a function to have a jump discontinuity?
4. How does changing the value of \( m \) affect the limits of the function?
5. Can you solve for \( m \) such that \( f(x) \) has a specific value at \( x = 1 \)?

**Tip:** Always check both one-sided limits at a point to confirm whether a function is continuous or has a discontinuity.

Let f be defined by a piecewise function where f(x) = 1 + mx if x ≤ 1, and f(x) = 4 - mx if x > 1. The questions involve graphing f for different values of m (m = 1, m = 2), calculating left-hand and right-hand limits as x approaches 1, finding m so that the left-hand and right-hand limits are equal, and describing how the graph changes for different values of m.

Explore the limits of a piecewise function defined by f(x) = 1 + mx for x ≤ 1 and f(x) = 4 - mx for x > 1. Solve for different values of m and analyze the continuity and discontinuity at x = 1. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation