Calculus Continuity on Closed Interval: Endpoint Considerations.

📚 Continuity on a Closed Interval: Endpoint Considerations

In calculus, understanding continuity on a closed interval requires special attention at the interval's endpoints. A function $f(x)$ is continuous on a closed interval $[a, b]$ if it is continuous at every point within the open interval $(a, b)$, and if it is also continuous from the right at $a$ and continuous from the left at $b$. Let's break this down further.

📜 A Brief History

The formalization of continuity evolved throughout the 19th century, building upon the intuitive understanding of curves and functions. Mathematicians like Cauchy and Weierstrass contributed significantly to our modern, rigorous definition, emphasizing the importance of limits and precise formulations like the epsilon-delta definition of continuity.

💡 Key Principles

• 🔍 Definition: A function $f(x)$ is continuous on the closed interval $[a, b]$ if it is continuous at every point in the open interval $(a, b)$, continuous from the right at $x = a$, and continuous from the left at $x = b$.
• ➡️ Right Continuity at $x = a$: The limit of $f(x)$ as $x$ approaches $a$ from the right must equal $f(a)$. Formally, $\lim_{x \to a^+} f(x) = f(a)$.
• ⬅️ Left Continuity at $x = b$: The limit of $f(x)$ as $x$ approaches $b$ from the left must equal $f(b)$. Formally, $\lim_{x \to b^-} f(x) = f(b)$.
• 📈 Continuity in the Open Interval $(a, b)$: For every $c$ in $(a, b)$, $\lim_{x \to c} f(x) = f(c)$.

➗ Examples

• 🌱 Example 1: Consider the function $f(x) = \sqrt{x}$ on the interval $[0, 1]$. At $x = 0$, we check the right-hand limit: $\lim_{x \to 0^+} \sqrt{x} = 0 = f(0)$. Thus, $f(x)$ is continuous from the right at $x = 0$. It is continuous for all $x > 0$, so $f(x)$ is continuous on $[0, 1]$.
• 🍂 Example 2: Let $g(x) = \begin{cases} x^2 & \text{if } 0 \leq x < 1 \\ 2-x & \text{if } 1 \leq x \leq 2 \end{cases}$ on the interval $[0, 2]$.
  • At $x = 0$, $\lim_{x \to 0^+} g(x) = 0 = g(0)$. Continuous from the right.
  • At $x = 1$, $\lim_{x \to 1^-} g(x) = 1$ and $\lim_{x \to 1^+} g(x) = 1$, and $g(1)=1$. Thus, $g(x)$ is continuous at $x = 1$.
  • At $x = 2$, $\lim_{x \to 2^-} g(x) = 0 = g(2)$. Continuous from the left.
  Therefore, $g(x)$ is continuous on $[0, 2]$.
• ❄️ Example 3: Consider the function $h(x) = \begin{cases} x & \text{if } 0 \leq x < 1 \\ 2 & \text{if } x = 1 \end{cases}$ on the interval $[0, 1]$. At $x = 1$, $\lim_{x \to 1^-} h(x) = 1 \neq h(1) = 2$. Thus, $h(x)$ is not continuous from the left at $x = 1$, and $h(x)$ is not continuous on $[0, 1]$.

📝 Practice Quiz

Determine the continuity of the following functions on the given closed intervals:

1. ❓ $f(x) = x^3 - 2x + 1$ on $[-1, 1]$
2. ❓ $g(x) = \frac{1}{x-2}$ on $[0, 1]$
3. ❓ $h(x) = \sqrt{4-x^2}$ on $[-2, 2]$
4. ❓ $k(x) = \begin{cases} x+1 & \text{if } -1 \leq x < 0 \\ x^2 & \text{if } 0 \leq x \leq 1 \end{cases}$ on $[-1, 1]$

🔑 Solutions

1. ✅ $f(x)$ is a polynomial and is continuous on $[-1, 1]$.
2. ❌ $g(x)$ has a discontinuity at $x = 2$, which is not in $[0, 1]$, but it is not continuous on $[0, 1]$ as it is discontinuous at x=2.
3. ✅ $h(x)$ is continuous on $[-2, 2]$.
4. ❌ $k(x)$ is not continuous at $x = 0$ because $\lim_{x \to 0^-} k(x) = 1$ and $\lim_{x \to 0^+} k(x) = 0$, so it is not continuous on $[-1, 1]$.

🎯 Conclusion

When evaluating continuity on closed intervals, always remember to check both endpoints for one-sided continuity in addition to checking continuity within the open interval. Mastering these endpoint considerations is key to accurately applying calculus concepts and theorems. Good luck! 👍