1.4 Q-2

Question Statement

Discuss the continuity of the given functions at the specified values of $x = c$:

1. $$$$

\begin{cases} 2x + 5 & \text{if } x \leq 2,
4x + 1 & \text{if } x > 2, \end{cases} \quad c = 2$$

2. $$$$

\begin{cases} 3x - 1 & \text{if } x < 1,
4 & \text{if } x = 1,
2x & \text{if } x > 1, \end{cases} \quad c = 1$$

3. $$$$

\begin{cases} 3x - 1 & \text{if } x < 1,
2x & \text{if } x > 1, \end{cases} \quad c = 1$$

Background and Explanation

A function $f(x)$ is continuous at $x = c$ if the following three conditions are met:

1. $f(c)$ is defined.
2. The left-hand limit ($\lim_{x \to c^-} f(x)$) and right-hand limit ($\lim_{x \to c^+} f(x)$) exist.
3. The left-hand limit, right-hand limit, and $f(c)$ are equal.

If any of these conditions fail, the function is discontinuous at $x = c$.

Solution

1. Function

\begin{cases} 2x + 5 & \text{if } x \leq 2, 4x + 1 & \text{if } x > 2, \end{cases} \quad c = 2$$ ### **Step 1: Evaluate $f(2)$** Substituting $x = 2$ into the first part of the function:

f(2) = 2(2) + 5 = 9

### **Step 2: Compute Left-Hand Limit (L.H.L)**

\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x + 5) = 2(2) + 5 = 9

### **Step 3: Compute Right-Hand Limit (R.H.L)**

\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (4x + 1) = 4(2) + 1 = 9

### **Step 4: Verify Continuity** Since $f(2) = 9$, $\text{L.H.L} = 9$, and $\text{R.H.L} = 9$, the function is continuous at $x = 2$. --- ## **2. Function** $$f(x) = \begin{cases} 3x - 1 & \text{if } x < 1, 4 & \text{if } x = 1, 2x & \text{if } x > 1, \end{cases} \quad c = 1$$ ### **Step 1: Evaluate $f(1)$** Substituting $x = 1$ from the second case:

f(1) = 4

### **Step 2: Compute Left-Hand Limit (L.H.L)**

\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3x - 1) = 3(1) - 1 = 2

### **Step 3: Compute Right-Hand Limit (R.H.L)**

\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x) = 2(1) = 2

### **Step 4: Verify Continuity** - $f(1) = 4$ - $\text{L.H.L} = 2$ - $\text{R.H.L} = 2$ Since $\text{L.H.L} \neq f(1)$ and $\text{R.H.L} \neq f(1)$, the function is **discontinuous** at $x = 1$. --- ## **3. Function** $$f(x) = \begin{cases} 3x - 1 & \text{if } x < 1, 2x & \text{if } x > 1, \end{cases} \quad c = 1$$ ### **Step 1: Evaluate $f(1)$** The function does not provide a definition for $f(1)$, so $f(1)$ is **not defined**. ### **Step 2: Verify Continuity** Since $f(1)$ is not defined, the function is **discontinuous** at $x = 1$. There is no need to compute the limits. --- # **Key Formulas or Methods Used** 1. **Definition of Continuity:** - $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$ 2. **Limit Calculation:** Substituting $x \to c^-$ and $x \to c^+$ into the respective cases of the function. --- # **Summary of Steps** 1. Evaluate $f(c)$ by substituting $x = c$ into the corresponding piece of the function. 2. Calculate the left-hand limit ($\lim_{x \to c^-} f(x)$). 3. Calculate the right-hand limit ($\lim_{x \to c^+} f(x)$). 4. Compare $f(c)$, L.H.L, and R.H.L: - If all are equal, the function is **continuous**. - If any are unequal or $f(c)$ is undefined, the function is **discontinuous**.