Piecewise-Defined Functions: Continuity and Differentiability Explained

Understanding Piecewise-Defined Functions 🧩

A piecewise-defined function is a function defined by multiple sub-functions, each applying to a certain interval of the main function's domain. For example:

f(x) = \begin{cases}
x^2, & x < 0 \\
x + 1, & 0 \le x \le 2 \\
3, & x > 2
\end{cases}

To determine continuity and differentiability, we need to analyze the "breakpoints" where the function definition changes (in this case, at $x = 0$ and $x = 2$).

Checking for Continuity 🔗

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

1. $f(a)$ is defined.
2. $\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$ both exist.
3. $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$.

Example: Checking Continuity at $x = 0$

• $f(0) = 0 + 1 = 1$ (defined)
• $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x^2 = 0$
• $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x + 1) = 1$

Since $\lim_{x \to 0^-} f(x) \ne \lim_{x \to 0^+} f(x)$, the function is not continuous at $x = 0$.

Example: Checking Continuity at $x = 2$

• $f(2) = 2 + 1 = 3$ (defined)
• $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x + 1) = 3$
• $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} 3 = 3$

Since $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$, the function is continuous at $x = 2$.

Checking for Differentiability 🔪

A function $f(x)$ is differentiable at $x = a$ if the following conditions are met:

1. $f(x)$ is continuous at $x = a$.
2. The left-hand derivative and the right-hand derivative exist at $x = a$.
3. The left-hand derivative equals the right-hand derivative at $x = a$.

The derivative is defined as:

f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

However, when dealing with piecewise functions, it's easier to find the derivatives of each piece and then check if they match at the breakpoints.

Example: Checking Differentiability at $x = 2$

First, we need to find the derivative of each piece:

f(x) = \begin{cases}
2x, & x < 0 \\
1, & 0 \le x \le 2 \\
0, & x > 2
\end{cases}

• Left-hand derivative at $x = 2$: $f'(2^-) = 1$
• Right-hand derivative at $x = 2$: $f'(2^+) = 0$

Since $f'(2^-) \ne f'(2^+)$, the function is not differentiable at $x = 2$, even though it's continuous there.

Key Takeaways 🚀

• Continuity is a prerequisite for differentiability. If a function is not continuous at a point, it cannot be differentiable at that point.
• To check for differentiability, ensure continuity first, then compare the left-hand and right-hand derivatives at the breakpoints.
• Piecewise functions require careful analysis at the points where their definitions change.