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Continuous functions are the building blocks of calculus. They behave predictably, allowing us to analyze their properties and apply them to real-world problems. Understanding these functions is crucial for grasping more complex mathematical concepts.

Properties of continuous functions, like the Intermediate Value Theorem, help us solve equations and model physical phenomena. We can also combine continuous functions through operations like addition and composition, creating more complex functions that retain continuity.

Properties of Continuous Functions

Sum, Difference, Product, and Quotient of Continuous Functions

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  • If f and g are continuous functions at a point a, then f + g, f - g, f * g, and f / g (provided g(a) ≠ 0) are also continuous at a
    • For example, if f(x) = x^2 and g(x) = sin(x) are continuous at x = π, then f(x) + g(x) = x^2 + sin(x) is also continuous at x = π
  • The sum, difference, product, and quotient of continuous functions are continuous on their respective domains
    • If f(x) = x^3 and g(x) = e^x are continuous on the real numbers, then f(x) - g(x) = x^3 - e^x is also continuous on the real numbers
  • These properties can be used to prove the continuity of more complex functions built from simpler continuous functions
    • The function h(x) = (x^2 + 1) / (x - 1) is continuous on its domain because the numerator and denominator are continuous functions, and the denominator is non-zero on the domain

Intermediate Value Theorem and Applications

  • The Intermediate Value Theorem (IVT) states that if f is continuous on a closed interval [a, b] and f(a) < k < f(b) or f(b) < k < f(a), then there exists a point c in (a, b) such that f(c) = k
    • If f(x) = x^3 - x is continuous on [-1, 1] and f(-1) = -2 < 0 < 2 = f(1), then there exists a point c in (-1, 1) such that f(c) = 0
  • The properties of continuous functions can be applied to solve problems involving limits, derivatives, and integrals
    • To find the limit of a continuous function f(x) as x approaches a, simply evaluate f(a)
    • The derivative of a continuous function is also continuous on its domain
    • Continuous functions on closed intervals are integrable, and the Fundamental Theorem of Calculus can be applied

Composition of Continuous Functions

Continuity of Composite Functions

  • If f is continuous at a and g is continuous at f(a), then the composite function g ∘ f is continuous at a
    • If f(x) = x^2 is continuous at x = 1 and g(x) = sin(x) is continuous at f(1) = 1, then g ∘ f(x) = sin(x^2) is continuous at x = 1
  • The composition of two continuous functions is continuous on its domain
    • If f(x) = x^3 and g(x) = e^x are continuous on the real numbers, then g ∘ f(x) = e^(x^3) is also continuous on the real numbers

Applications of Composite Function Continuity

  • The continuity of composite functions can be used to prove the continuity of more complex functions
    • The function h(x) = sin(cos(x)) is continuous on the real numbers because sin(x) and cos(x) are continuous functions
  • The chain rule for derivatives and the substitution rule for integrals rely on the continuity of composite functions
    • If f(x) and g(x) are differentiable functions, then the derivative of the composite function g ∘ f(x) is given by (gf)(x)=g(f(x))f(x)(g ∘ f)'(x) = g'(f(x)) · f'(x)
    • When evaluating the definite integral abf(g(x))g(x)dx\int_a^b f(g(x))g'(x)dx, the substitution rule can be applied by setting u=g(x)u = g(x) and du=g(x)dxdu = g'(x)dx

Continuity of Piecewise Functions

Conditions for Continuity of Piecewise Functions

  • A piecewise-defined function is continuous if each piece is continuous on its respective domain and the function values agree at the endpoints of adjacent pieces
    • The absolute value function f(x)=x={x,x0x,x<0f(x) = |x| = \begin{cases} x, & x \geq 0 \\ -x, & x < 0 \end{cases} is continuous because both pieces are continuous on their respective domains, and the function values agree at x = 0
  • To determine the continuity of a piecewise-defined function, check the continuity of each piece and the agreement of function values at the endpoints
    • For the function f(x)={x2,x12x1,x>1f(x) = \begin{cases} x^2, & x \leq 1 \\ 2x - 1, & x > 1 \end{cases}, check the continuity of x^2 on (-∞, 1] and 2x - 1 on (1, ∞), and verify that the function values agree at x = 1

Types of Discontinuities in Piecewise Functions

  • Piecewise-defined functions can have removable, jump, or infinite discontinuities at the endpoints of the pieces
    • The function f(x)={x21x1,x12,x=1f(x) = \begin{cases} \frac{x^2 - 1}{x - 1}, & x \neq 1 \\ 2, & x = 1 \end{cases} has a removable discontinuity at x = 1
    • The function f(x)={0,x<01,x0f(x) = \begin{cases} 0, & x < 0 \\ 1, & x \geq 0 \end{cases} has a jump discontinuity at x = 0
    • The function f(x)={1x,x>00,x0f(x) = \begin{cases} \frac{1}{x}, & x > 0 \\ 0, & x \leq 0 \end{cases} has an infinite discontinuity at x = 0
  • Removable discontinuities can be eliminated by redefining the function value at the point of discontinuity to make the function continuous
    • For the function f(x)={x21x1,x12,x=1f(x) = \begin{cases} \frac{x^2 - 1}{x - 1}, & x \neq 1 \\ 2, & x = 1 \end{cases}, redefining f(1) = 2 makes the function continuous at x = 1
  • Jump discontinuities occur when the left-hand and right-hand limits at a point exist but are not equal
    • For the function f(x)={0,x<01,x0f(x) = \begin{cases} 0, & x < 0 \\ 1, & x \geq 0 \end{cases}, limx0f(x)=0\lim_{x \to 0^-} f(x) = 0 and limx0+f(x)=1\lim_{x \to 0^+} f(x) = 1, resulting in a jump discontinuity at x = 0
  • Infinite discontinuities occur when either the left-hand or right-hand limit at a point is infinite
    • For the function f(x)={1x,x>00,x0f(x) = \begin{cases} \frac{1}{x}, & x > 0 \\ 0, & x \leq 0 \end{cases}, limx0+f(x)=\lim_{x \to 0^+} f(x) = \infty, resulting in an infinite discontinuity at x = 0


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© 2025 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
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