3.2 Properties of continuous functions

Continuous functions have special properties that make them powerful tools in calculus. These properties allow us to manipulate and analyze functions with ease, opening doors to solving complex problems.

Understanding how continuous functions behave when added, multiplied, or composed is crucial. These properties, along with theorems like the Extreme Value Theorem, help us prove important mathematical statements and solve real-world problems.

Properties of Continuous Functions

Properties of continuous functions

• Sum of continuous functions remains continuous at the same point $x=a$
  • Proven using limit laws: $\lim_{x \to a} (f(x)+g(x)) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) = f(a) + g(a)$
• Difference of continuous functions remains continuous at the same point $x=a$
  • Proven using limit laws: $\lim_{x \to a} (f(x)-g(x)) = \lim_{x \to a} f(x) - \lim_{x \to a} g(x) = f(a) - g(a)$
• Product of continuous functions remains continuous at the same point $x=a$
  • Proven using limit laws: $\lim_{x \to a} (f(x)g(x)) = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) = f(a)g(a)$
• Quotient of continuous functions remains continuous at the same point $x=a$, provided the denominator is non-zero at $x=a$
  • Proven using limit laws: $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} = \frac{f(a)}{g(a)}$, where $g(a) \neq 0$

Continuity of composite functions

• Composite function $g(f(x))$ is continuous at $x=a$ if:
  • Inner function $f(x)$ is continuous at $x=a$
  • Outer function $g(x)$ is continuous at $x=f(a)$
• Proven using limit laws: $\lim_{x \to a} g(f(x)) = g(\lim_{x \to a} f(x)) = g(f(a))$
• Determine continuity by checking both conditions for the inner and outer functions
  • Example: If $f(x) = x^2$ is continuous on $\mathbb{R}$ and $g(x) = \sqrt{x}$ is continuous on $[0, \infty)$, then $g(f(x)) = \sqrt{x^2} = |x|$ is continuous on $\mathbb{R}$

Extreme value theorem

• Continuous function $f(x)$ on a closed interval $[a,b]$ attains its maximum and minimum values within the interval
• Proven by contradiction:
  1. Assume $f(x)$ does not attain its maximum value on $[a,b]$
  2. Let $M = \sup\{f(x) : x \in [a,b]\}$ be the least upper bound of $f(x)$ on $[a,b]$
  3. For each $n \in \mathbb{N}$, there exists $x_n \in [a,b]$ such that $f(x_n) > M - \frac{1}{n}$
  4. By the Bolzano-Weierstrass Theorem, the sequence $(x_n)$ has a convergent subsequence $(x_{n_k})$ converging to some $x_0 \in [a,b]$
  5. By continuity, $\lim_{k \to \infty} f(x_{n_k}) = f(x_0)$, but $\lim_{k \to \infty} f(x_{n_k}) = M$, so $f(x_0) = M$, contradicting the assumption
  6. A similar proof can be used for the minimum value
• Ensures the existence of maximum and minimum values for continuous functions on closed intervals

Applications of function continuity

• Intermediate Value Theorem: If $f(x)$ is continuous on $[a,b]$ and $y_0$ is between $f(a)$ and $f(b)$, then there exists a $c \in (a,b)$ such that $f(c) = y_0$
  • Proves the existence of roots and helps find the range of a function
  • Example: If $f(x) = x^3 - x$ is continuous on $[-1,1]$ and $f(-1) = -2 < 0$ and $f(1) = 0$, then there exists a $c \in (-1,1)$ such that $f(c) = -1$
• Boundedness Theorem: If $f(x)$ is continuous on a closed interval $[a,b]$, then $f(x)$ is bounded on $[a,b]$
  • Proves a function is bounded and helps find the range of a function
  • Example: If $f(x) = \sin(x)$ is continuous on $[0, 2\pi]$, then $f(x)$ is bounded on $[0, 2\pi]$ with $-1 \leq f(x) \leq 1$
• Preservation of intervals: If $f(x)$ is continuous on an interval $I$ and $f(I) \subseteq J$, then for any interval $K \subseteq J$, there exists an interval $L \subseteq I$ such that $f(L) = K$
  • Proves the existence of pre-images and helps find the domain of a composite function
  • Example: If $f(x) = x^2$ is continuous on $\mathbb{R}$ and $f(\mathbb{R}) = [0, \infty)$, then for any interval $[a,b] \subseteq [0, \infty)$, there exists an interval $[-\sqrt{b}, -\sqrt{a}] \cup [\sqrt{a}, \sqrt{b}] \subseteq \mathbb{R}$ such that $f([-\sqrt{b}, -\sqrt{a}] \cup [\sqrt{a}, \sqrt{b}]) = [a,b]$

Applying Continuity Properties

Applying properties to solve problems and prove statements

• Solving equations using the Intermediate Value Theorem
  • Proves the existence of a solution for a continuous function $f(x) = 0$ on an interval $[a,b]$ if $f(a)$ and $f(b)$ have opposite signs
  • Example: If $f(x) = x^3 - x - 1$ is continuous on $[0,2]$ and $f(0) = -1 < 0$ and $f(2) = 5 > 0$, then there exists a $c \in (0,2)$ such that $f(c) = 0$
• Proving inequalities using the Extreme Value Theorem
  • If $f(x)$ and $g(x)$ are continuous on $[a,b]$ and $f(x) \leq g(x)$ for all $x \in [a,b]$, then $\max f(x) \leq \max g(x)$ and $\min f(x) \leq \min g(x)$
  • Example: If $f(x) = \sin(x)$ and $g(x) = \cos(x)$ are continuous on $[0, \frac{\pi}{2}]$, then $\max f(x) = 1 \leq \max g(x) = 1$ and $\min f(x) = 0 \leq \min g(x) = \frac{\sqrt{2}}{2}$
• Approximating solutions using the Intermediate Value Theorem and the Bisection Method
  • If $f(x)$ is continuous and $f(a)$ and $f(b)$ have opposite signs, the Bisection Method can approximate a solution to $f(x) = 0$ with arbitrary precision
  • Example: If $f(x) = x^3 - x - 1$ is continuous on $[1,2]$ and $f(1) = -1 < 0$ and $f(2) = 5 > 0$, the Bisection Method can approximate a solution to $f(x) = 0$ by repeatedly halving the interval and selecting the subinterval where the function changes sign