Continuity

Section 1.4 Continuity

A continuous graph can be thought of as a graph that you can draw without lifting your pencil.

Here are some examples of discontinuities:

Figure 1.4.1. Removeable Discontinuity

This is an example of a removeable discontinuity. These can have or not have a point, and it simply a hole on an otherwise continuous graph.
Figure 1.4.2. Jump Discontinuity

This is a jump discontinuity. Pretty self explanitory.
Figure 1.4.3. Infinite Discontinuity

Finally, this is an infinite discontinuity. These can be describes as having an asymptotes in one or more directions.

Definition 1.4.4. Continuity.

A given function \(f(x)\) is said to be continuous as a point \(x = c\) provided:

\begin{equation*} \lim_{x \to c} f(x) = f(c) \end{equation*}

We say a function is continuous on the open interval \((a \mbox{ , } b)\) if it is continuous at all point inside the interval.

We say that a function is continuous everywhere if it is continuous at all point on the interval \((- \infty \mbox{ , } \infty)\text{.}\)

Theorem 1.4.5. Continuity.

a polynomial is continuous everywhere
a rational function is continuous at every point where the denominator is not equal to zero and discontinuous where the denominator equals zero

For the function \(P_{n}(x) = C_0 + C_1x^1 + \dots + C_nx^n\text{,}\) to find \(\displaystyle\lim_{x \to c} P_n(x)\text{,}\) we would take the limit of the formula, reducing to:

\begin{equation*} \lim_{x \to c} P_n(x) = P_n(c) \end{equation*}

Theorem 1.4.6. Properties of Continuity.

If \(f(x)\) and \(g(x)\) are continuous at \(x = c\text{:}\)

\(f(x) + g(x)\) is continuous at \(x = c\)
\(f(x) - g(x)\) is continuous at \(x = c\)
\(f(x) \cdot g(x)\) is continuous at \(x = c\)
\(f(x) / g(x)\) is continuous at \(x = c\) provided \(g(x) \neq 0\)

(if \(g(x) = 0\) then \(f(x) / g(x)\) has a disconinuity)

Theorem 1.4.7. Intermediate Value Theorem.

If \(f(x)\) is continuous on a closed interval, \([ a \mbox{ , } b]\text{,}\) and \(k\) is any number at or between \(f(a) \mbox{ and } f(b)\text{,}\) then there is at least one number, \(x\text{,}\) in \([ a \mbox{ , } b]\) such that \(f(x) = k\text{.}\)

Figure 1.4.8.

Figure 1.4.9.

\(f(x)\) isn't continuous on \([ a \mbox{ , } b]\) because \(f(a) \neq \displaystyle\lim_{x \to a^+} f(x)\)

\(f(x)\) is continuous on \([ a \mbox{ , } b]\) because \(f(a) = \displaystyle\lim_{x \to a^+} f(x)\) and \(f(b) = \displaystyle\lim_{x \to b^-} f(x)\)

Idea of I.V.T..

Figure 1.4.10.

a spectial value of \(k\) we often are interested in is when \(k = 0\)
when a function is continuous, it means that we can't just "hop over" the values that are intercepts

Theorem 1.4.11. Continuity Solutions.

If \(f(x)\) is continuous on the closed interval \([ a \mbox{ , } b ]\) and if \(f(a)\) and \(f(b)\) are both nonzero and have opposite signs, then there is at least one solution to \(f(x) = 0\) on the interval \([ a \mbox{ , } b ]\text{.}\) There may be more than one answer.

Figure 1.4.12.

Example 1.4.1.

Show that \(x^3 - x = 1\) has a solutions in \([1\mbox{ , }2]\text{:}\)

Let \(f(x) = x^3 - x - 1\)

we have:
- function
- interval
we don't have:
- function values
  
  (we can find them as shown to the right)

\begin{equation*} f(1) = 1^3 - 1 - 1 = -1 (\lt 0) \end{equation*}

\begin{equation*} f(2) = 2^3 - 2 - 1 = 5 (\gt 0) \end{equation*}

Since \(f(x)\) is continuous and \(f(1)\) and \(f(2)\) are opposite in sign, \(f(x)\) has at least one root in \(( 1 \mbox{ , } 2 )\text{;}\) this means \(x^3 - x = 1\) also has at least one solution in the interval.

We haven't found an exact solution but we now know that it is possible.

To summarize, continuous functions have nive properties. We can replace limit problems with function evaluation problems when the given function is continuous at the point where we are talking the limit.