Continuity & Differentiability
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Part B: Differentiability Exercises for This Topic Index of On-Line Topics Everything for Calculus Everything for Finite Math Everything for Finite Math & Calculus Utility: Function Evaluator & Grapher Español |
Part A: Continuity
Note To understand this topic, you will need to be familiar with limits, as discussed in the chapter on derivatives in Calculus Applied to the Real World. If you like, you can review the topic summary material on limits or, for a more detailed study, the on-line tutorial on limits.
To begin, we recall the definition (see Section 6 in the derivatives chapter in Calculus Applied to the Real World.) of what it means for a function to be continuous at a point or on a subset of its domain.
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Continuous at a Point
The function $f$ is continuous at the point a in its domain if:
If the point $a$ is not in the domain of $f,$ we do not talk about whether or not $f$ is continuous at $a.$ Continuous on a Subset of the Domain The function $f$ is continuous on the subset $S$ of its domain if it continuous at each point of $S.$ Examples 1. All closed form functions are continuous on their (whole) domain. A closed-form function is any function that can be obtained by combining constants, powers of $x,$ exponential functions, radicals, logarithms, and trigonometric functions (and some other functions we shall not encounter here) into a single mathematical formula by means of the usual arithmetic operations and composition of functions. Examples of closed form functions are:
2. The function $f(x) = 1/x,$ also of closed form, is continuous at every point of its domain. (Note that 0 is not a point of the domain of $f,$ so we don't discuss what it might mean to be continous or discontinuous there.) 3. The function
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Example 1 Recognizing Points of Discontinuity in a Graph
Let f have the graph shown below.
Solution
According to the definition, $f$ can fail to be continuous at a point $a$ of its domain if either:| 1. | $lim$ $x→a$ | $f(x)$ does not exist, or |
| 2. | $lim$ $x→a$ | $f(x)$ exists, but $≠ f(a)$ |
$x = 0:$ From the graph, we see that $f(0)$ is not defined. Therefore, $0$ is not in the domain of $f,$ so we cannot say that $f$ has a discontinuity at $x = 0.$ (Some authors would say that $f$ is discontinuous at $x = 0,$ but we don't consider such points...)
$x = 1:$ At the point where $x = 1,$ the limit does exist, however,
| $lim$ $x→1$ | $f(x) = 1,$ but |
| $f(1) = 2$ |
Here is a more interactive example.
Example 2 Recognizing Points of Discontinuity in a Graph
The next examples discuss functions specified algebraically rather than geometrically.
Example 3 Identifying Points of Discontinuity in a Non-Closed Form Function
| $f(x) =$ |  | $x^2+4$ $x + 1$ $x^2 + 1$ | if $x ≤ 0$ if $0 < x ≤ 1$ if $x > 1$ |
$.$ |
| $lim$ $x→0^{-}$ | $f(x) = 4$ | Substitute $x = 0$ in the first (closed-form) formula |
| $lim$ $x→0^{+}$ | $f(x) = 1$ | Substitute $x = 0$ in the second (closed-form) formula |
Example 4 Fixing up a Function
| $f(x) =$ | | $x^2 - x + 1$ $kx^2 + 1$ | if $x ≤ 1$ if $x > 1$ |
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Solution
Since the only place where something can go wrong is at $x = 1,$ we look at the left- and right limits, as well as the value of the function.| $lim$ $x→1^{-}$ | $f(x) = 1$ |
| $lim$ $x→1^{+}$ | $f(x) = k+1$ |
| $f(1) = 1$ | |
You can now either go on and try those exercises that deal with continuity in the exercise set for this topic, or first finish the on-line text by going to Part B: Differentiability.