Continuity, Differentiability and Mean Value Theorems

1. Continuity of Functions

In mathematics, a function is said to be continuous if its graph can be drawn without lifting the pen from the paper. More formally, a function $f(x)$ is continuous at a point $c$ in its domain if three conditions are met:

1. The function is defined at $c$, meaning $f(c)$ exists.
2. The limit of the function as $x$ approaches $c$ exists, meaning $\lim_{x \to c} f(x)$ exists.
3. The limit of the function as $x$ approaches $c$ is equal to the function's value at $c$, meaning $\lim_{x \to c} f(x) = f(c)$.

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

Types of Discontinuities

There are several types of discontinuities:

• Removable Discontinuity: Occurs when $\lim_{x \to c} f(x)$ exists, but either $f(c)$ is undefined or $\lim_{x \to c} f(x) \neq f(c)$. This type of discontinuity can be "removed" by redefining the function at $c$.
• Jump Discontinuity: Occurs when the left-hand limit ($\lim_{x \to c^-} f(x)$) and the right-hand limit ($\lim_{x \to c^+} f(x)$) both exist, but are not equal.
• Infinite Discontinuity: Occurs when at least one of the one-sided limits is infinite (either $+\infty$ or $-\infty$). This often happens when the function has a vertical asymptote at $c$.
• Oscillating Discontinuity: Occurs when the function oscillates infinitely many times between two values as $x$ approaches $c$, so the limit does not exist.

Continuity on an Interval

A function $f$ is continuous on an open interval $(a, b)$ if it is continuous at every point in the interval.

A function $f$ is continuous on a closed interval $[a, b]$ if it is continuous on $(a, b)$, and the right-hand limit at $a$ equals $f(a)$ ($\lim_{x \to a^+} f(x) = f(a)$), and the left-hand limit at $b$ equals $f(b)$ ($\lim_{x \to b^-} f(x) = f(b)$).

Properties of Continuous Functions

If $f$ and $g$ are continuous functions at $c$, then the following functions are also continuous at $c$:

• $f+g$ (Sum): $(f+g)(x) = f(x) + g(x)$
• $f-g$ (Difference): $(f-g)(x) = f(x) - g(x)$
• $f \cdot g$ (Product): $(f \cdot g)(x) = f(x) \cdot g(x)$
• $\frac{f}{g}$ (Quotient): $(\frac{f}{g})(x) = \frac{f(x)}{g(x)}$, provided $g(c) \neq 0$.
• $f \circ g$ (Composition): $(f \circ g)(x) = f(g(x))$, provided $f$ is continuous at $g(c)$.

Polynomials, rational functions (where the denominator is non-zero), trigonometric functions (sin, cos), exponential functions, and logarithmic functions are continuous on their respective domains.

2. Differentiability of Functions

A function $f(x)$ is differentiable at a point $c$ if its derivative exists at $c$. The derivative of a function at a point represents the instantaneous rate of change of the function, or the slope of the tangent line to the function's graph at that point.

The derivative of $f$ at $c$, denoted by $f'(c)$, is defined by the limit:

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

This limit must exist for the function to be differentiable at $c$. This definition is equivalent to:

$$ f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x-c} $$

For the derivative to exist, the limit must be the same whether $x$ approaches $c$ from the left or the right. This means the left-hand derivative must equal the right-hand derivative:

$$ \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h} = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h} $$

Relationship between Continuity and Differentiability

A crucial theorem states: If a function $f$ is differentiable at a point $c$, then it must be continuous at $c$.

However, the converse is not true: A function can be continuous at a point but not differentiable at that point.

A common example of a function that is continuous but not differentiable at a point is the absolute value function, $f(x) = |x|$, at $x=0$. The graph has a sharp corner at the origin, meaning the slope from the left is -1 and the slope from the right is +1. Since these do not match, the derivative does not exist at $x=0$, even though the function is continuous there.

Where Differentiability Fails

A function may fail to be differentiable at a point $c$ due to:

• Discontinuity: If $f$ is not continuous at $c$, it cannot be differentiable at $c$.
• Corner or Cusp: The left-hand and right-hand derivatives exist but are unequal (e.g., $f(x) = |x|$ at $x=0$).
• Vertical Tangent Line: The limit defining the derivative approaches $\infty$ or $-\infty$. For example, $f(x) = \sqrt[3]{x}$ at $x=0$ has a vertical tangent.

Differentiation Rules

These rules allow us to find derivatives of complex functions by combining simpler ones. Let $u$ and $v$ be differentiable functions of $x$.

• Constant Rule: $\frac{d}{dx}(c) = 0$, where $c$ is a constant.
• Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$.
• Constant Multiple Rule: $\frac{d}{dx}(c \cdot f(x)) = c \cdot \frac{d}{dx}(f(x))$.
• Sum/Difference Rule: $\frac{d}{dx}(u \pm v) = \frac{du}{dx} \pm \frac{dv}{dx}$.
• Product Rule: $\frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx}$.
• Quotient Rule: $\frac{d}{dx}(\frac{u}{v}) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$, provided $v \neq 0$.
• Chain Rule: $\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)$. If $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

3. Mean Value Theorems

Mean Value Theorems are fundamental results in calculus that relate the values of a function to its derivative. They provide important guarantees about the behavior of differentiable functions.

3.1 Rolle's Theorem

Statement: Let $f$ be a function that satisfies the following three conditions:

1. $f$ is continuous on the closed interval $[a, b]$.
2. $f$ is differentiable on the open interval $(a, b)$.
3. $f(a) = f(b)$.

Then, there exists at least one number $c$ in the open interval $(a, b)$ such that $f'(c) = 0$.

Geometric Interpretation: If a curve is smooth (continuous and differentiable) and starts and ends at the same height ($f(a)=f(b)$), then somewhere between the endpoints, there must be at least one point where the tangent line is horizontal ($f'(c)=0$).

Example: Consider the function $f(x) = x^2 - 4x + 3$ on the interval $[1, 3]$.

• $f(x)$ is a polynomial, so it's continuous on $[1, 3]$ and differentiable on $(1, 3)$.
• $f(1) = 1^2 - 4(1) + 3 = 1 - 4 + 3 = 0$.
• $f(3) = 3^2 - 4(3) + 3 = 9 - 12 + 3 = 0$.
• Since $f(1) = f(3)$, Rolle's Theorem applies.

We need to find $c$ such that $f'(c) = 0$. The derivative is $f'(x) = 2x - 4$. Setting $f'(c) = 0$, we get $2c - 4 = 0$, which gives $c = 2$. Since $c=2$ is in the interval $(1, 3)$, Rolle's Theorem is satisfied.

3.2 Mean Value Theorem (MVT)

Statement: Let $f$ be a function that satisfies the following two conditions:

1. $f$ is continuous on the closed interval $[a, b]$.
2. $f$ is differentiable on the open interval $(a, b)$.

Then, there exists at least one number $c$ in the open interval $(a, b)$ such that:

$$ f'(c) = \frac{f(b) - f(a)}{b - a} $$

Geometric Interpretation: If a curve is smooth between two points $(a, f(a))$ and $(b, f(b))$, then there must be at least one point $c$ between $a$ and $b$ where the slope of the tangent line ($f'(c)$) is equal to the slope of the secant line connecting the two endpoints ($\frac{f(b) - f(a)}{b - a}$).

Relationship to Rolle's Theorem: Rolle's Theorem is a special case of the Mean Value Theorem. If $f(a) = f(b)$, then $\frac{f(b) - f(a)}{b - a} = \frac{0}{b - a} = 0$. The MVT then states that there exists a $c$ such that $f'(c) = 0$, which is exactly Rolle's Theorem.

Example: Consider the function $f(x) = x^3$ on the interval $[0, 2]$.

• $f(x)$ is a polynomial, so it's continuous on $[0, 2]$ and differentiable on $(0, 2)$.
• $f(0) = 0^3 = 0$.
• $f(2) = 2^3 = 8$.
• The slope of the secant line is $\frac{f(2) - f(0)}{2 - 0} = \frac{8 - 0}{2} = 4$.

The MVT guarantees there exists a $c$ in $(0, 2)$ such that $f'(c) = 4$. The derivative is $f'(x) = 3x^2$. Setting $f'(c) = 4$, we get $3c^2 = 4$, so $c^2 = \frac{4}{3}$, and $c = \pm \frac{2}{\sqrt{3}}$. Since we are looking for $c$ in $(0, 2)$, we choose $c = \frac{2}{\sqrt{3}}$. This value is indeed in $(0, 2)$.

3.3 Cauchy's Mean Value Theorem (Generalized Mean Value Theorem)

Statement: Let $f$ and $g$ be two functions that satisfy the following conditions:

1. $f$ and $g$ are continuous on the closed interval $[a, b]$.
2. $f$ and $g$ are differentiable on the open interval $(a, b)$.
3. $g'(x) \neq 0$ for all $x$ in $(a, b)$.

Then, there exists at least one number $c$ in the open interval $(a, b)$ such that:

$$ \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)} $$

Relationship to MVT: If we let $g(x) = x$, then $g'(x) = 1$. $g(x)$ is continuous and differentiable everywhere, and $g'(x) \neq 0$. In this case, Cauchy's MVT becomes:

$$ \frac{f'(c)}{1} = \frac{f(b) - f(a)}{b - a} $$

This is precisely the standard Mean Value Theorem. Cauchy's MVT is a generalization that deals with the ratio of derivatives.

Application: L'Hôpital's Rule

Cauchy's Mean Value Theorem is a key step in proving L'Hôpital's Rule, which is used to evaluate indeterminate forms of limits (like $\frac{0}{0}$ or $\frac{\infty}{\infty}$).

If $\lim_{x \to c} \frac{f(x)}{g(x)}$ results in an indeterminate form, and $f$ and $g$ satisfy the conditions of Cauchy's MVT near $c$, then:

$$ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} $$

This rule allows us to replace the limit of a ratio of functions with the limit of the ratio of their derivatives, which is often easier to compute.

3.4 Lagrange's Mean Value Theorem

Lagrange's Mean Value Theorem is simply another name for the standard Mean Value Theorem (MVT). It emphasizes the existence of a point $c$ where the instantaneous rate of change equals the average rate of change.

4. Applications of Derivatives and Mean Value Theorems

Continuity and differentiability are foundational concepts, and the Mean Value Theorems connect these concepts to the behavior of functions. Derivatives themselves have numerous applications:

• Optimization: Finding maximum and minimum values of functions (using critical points where $f'(x)=0$ or $f'(x)$ is undefined, and checking endpoints).
• Curve Sketching: Determining intervals of increase/decrease (from the sign of $f'(x)$) and concavity (from the sign of $f''(x)$).
• Related Rates: Solving problems where multiple quantities change over time and their rates of change are related.
• Approximation: Using tangent lines (linear approximation) to estimate function values near a known point.

The Mean Value Theorems provide theoretical underpinnings for many of these applications. For instance, they guarantee that if a function's derivative is always positive on an interval, the function must be strictly increasing on that interval.

Example: Using MVT to show a function is increasing

Suppose $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, and $f'(x) > 0$ for all $x$ in $(a, b)$. We want to show $f$ is increasing on $[a, b]$.

Let $x_1, x_2$ be any two numbers in $[a, b]$ such that $x_1 < x_2$. Consider the interval $[x_1, x_2]$.

By the MVT applied to $f$ on $[x_1, x_2]$, there exists a $c$ in $(x_1, x_2)$ such that:

$$ f'(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1} $$

Since $f'(x) > 0$ for all $x$ in $(a, b)$, we know $f'(c) > 0$.

Therefore, $\frac{f(x_2) - f(x_1)}{x_2 - x_1} > 0$.

Since the denominator $(x_2 - x_1)$ is positive (because $x_1 < x_2$), the numerator must also be positive: $f(x_2) - f(x_1) > 0$.

This implies $f(x_2) > f(x_1)$. Since this holds for any $x_1 < x_2$ in $[a, b]$, the function $f$ is strictly increasing on $[a, b]$.