- Outline the three criteria a function must meet to be continuous at a specific point
- Explain the different types of breaks a function can have that make it not continuous
- Explain what it means for a function to be continuous over a range of values
- Explain the rule for calculating limits of functions that are combined
- Show how a continuous function reaches every value between its start and end points using the Intermediate Value Theorem
Continuity
The Main Idea
- A function [latex]f(x)[/latex] is continuous at [latex]x = a[/latex] if:
- f(a) is defined
- [latex]\lim_{x \to a} f(x)[/latex] exists
- [latex]\lim_{x \to a} f(x) = f(a)[/latex]
- Types of Discontinuities:
- Removable (hole in the graph)
- Jump (function value jumps at a point)
- Infinite (function approaches infinity near a point)
- Continuity of Functions:
- Polynomials are continuous everywhere
- Rational functions are continuous except where denominator is zero
- Analyzing Continuity:
- Check if function is defined at the point
- Evaluate the limit as [latex]x[/latex] approaches the point
- Compare the limit value with the function value at the point
Using the definition, determine whether the function [latex]f(x)=\begin{cases} 2x+1 & \text{ if } \, x < 1 \\ 2 & \text{ if } \, x = 1 \\ -x+4 & \text{ if } \, x > 1 \end{cases}[/latex] is continuous at [latex]x=1[/latex].
If the function is not continuous at [latex]1[/latex], indicate the condition for continuity at a point that fails to hold.
For what values of [latex]x[/latex] is [latex]f(x)=3x^4-4x^2[/latex] continuous?
Types of Discontinuities
The Main Idea
- Three Main Types of Discontinuities:
- Removable (hole in the graph)
- Jump (function value jumps at a point)
- Infinite (function approaches infinity near a point)
- Removable Discontinuity:
- Limit exists at the point, but function value is different or undefined
- Graphically appears as a hole in the function
- Jump Discontinuity:
- Left-hand and right-hand limits exist but are not equal
- Function “jumps” from one value to another
- Infinite Discontinuity:
- Function approaches infinity as it nears the point
- Often associated with vertical asymptotes
- Identifying Discontinuities:
- Evaluate function at the point
- Calculate left-hand and right-hand limits
- Compare limits to function value
For [latex]f(x)=\begin{cases} x^2 & \text{ if } \, x \ne 1 \\ 3 & \text{ if } \, x = 1 \end{cases}[/latex], decide whether [latex]f[/latex] is continuous at [latex]1[/latex].
If [latex]f[/latex] is not continuous at [latex]1[/latex], classify the discontinuity as removable, jump, or infinite.
State the interval(s) over which the function [latex]f(x)=\sqrt{x+3}[/latex] is continuous.
Continuity Over an Interval
The Main Idea
- Continuity on Open Intervals:
- Function is continuous at every point within [latex](a, b)[/latex]
- Continuity on Closed Intervals:
- Function is continuous on [latex](a, b)[/latex]
- Right-continuous at [latex]a[/latex]: [latex]\lim_{x \to a^+} f(x) = f(a)[/latex]
- Left-continuous at [latex]b[/latex]: [latex]\lim_{x \to b^-} f(x) = f(b)[/latex]
- Half-Open Intervals:
- For [latex](a, b][/latex]: Continuous on [latex](a, b)[/latex] and left-continuous at [latex]b[/latex]
- For [a, b): Continuous on [latex](a, b)[/latex] and right-continuous at [latex]a[/latex]
- Determining Continuity:
- Check domain of the function
- Analyze behavior at endpoint(s) for closed intervals
- Consider discontinuities within the interval
Determine the interval(s) of continuity for the function:
[latex]f(x) = \begin{cases} \sqrt{x+1} & \text{if } x < 3 \\ \frac{x-3}{x-2} & \text{if } x \geq 3 \end{cases}[/latex]
Composite Function Theorem and The Intermediate Value Theorem
The Main Idea
- Composite Function Theorem:
- If [latex]f(x)[/latex] is continuous at [latex]L[/latex] and [latex]\lim_{x \to a} g(x) = L[/latex], then: [latex]\lim_{x \to a} f(g(x)) = f(\lim_{x \to a} g(x)) = f(L)[/latex]
- Helps expand our ability to compute limits
- Demonstrates continuity of trigonometric functions
- Continuity of Trigonometric Functions:
- All trigonometric functions are continuous over their entire domains
- Proof uses Composite Function Theorem and continuity of [latex]\sin x[/latex] and [latex]\cos x[/latex] at [latex]0[/latex]
- Intermediate Value Theorem (IVT):
- Applies to functions continuous over closed, bounded intervals [latex][a,b][/latex]
- If [latex]z[/latex] is between [latex]f(a)[/latex] and [latex]f(b)[/latex], there exists [latex]c[/latex] in [latex][a,b][/latex] where [latex]f(c) = z[/latex]
- Useful for proving existence of solutions (e.g., zeros of functions)
- Applications of IVT:
- Proving existence of zeros for continuous functions
- Cannot be used to prove non-existence of zeros or other values
For [latex]f(x)=1/x, \, f(-1)=-1<0[/latex] and [latex]f(1)=1>0[/latex]. Can we conclude that [latex]f(x)[/latex] has a zero in the interval [latex][-1,1][/latex]?