Wemazi

Topic 1.13

Define a continuous extension

Builds on finding the missing value. 7 minutes.

A dot that must fit

The top of \(f(x) = \dfrac{x^2 - 2x - 3}{x + 1}\) factors as \((x - 3)(x + 1)\), so the graph is the line \(y = x - 3\) with a ring at \(x = -1\). At \(-1\) the formula gives 0/0, so \(f\) has no value there.

Move the orange dot along the faint line \(x = -1\). Each height is a possible value at \(-1\).

The red segment shows the gap between the dot and the ring, and it closes only at the height \(-4\).

The height \(-4\) is the limit of \(f\) at \(-1\), and the new value must equal it so that the graph has no break.

Fill one point and keep the rest

The curve \(f\) has a ring at \(c\), so \(f(c)\) is not defined.

Press the button. The new function \(F\) has an orange dot in the ring, at the height \(L\) that the curve approaches.

At every input except \(c\), \(F\) has the same value as \(f\), so only the point at \(c\) is new.

The graph of \(F\) has no break, so \(F\) is continuous at \(c\). This \(F\) is the continuous extension of \(f\).

Notation

  1. A continuous extension has two rules. Away from \(c\), it copies \(f\). At \(c\), it takes the value \(L\) of the limit.

    \[\begin{aligned} F(x) &= \begin{cases} f(x) & x \neq c \\ \cy{L} & x = c \end{cases} \\ L &= \lim_{x\to c} f(x) \end{aligned}\]

  2. The limit must exist and be a real number. If it does not, there is no value \(L\) to use.

  3. The function \(F\) is continuous at \(c\). A limit ignores the point at \(c\), and \(F\) equals \(f\) near \(c\), so the limit of \(F\) is also \(L\). The value \(F(c)\) is \(L\) by the definition.

    \[\lim_{x\to c} F(x) = L = F(c)\]

  4. If \(f(x) = g(x)\) for \(x \neq c\) and \(g\) is continuous at \(c\), then \(L = g(c)\), so \(F = g\) at every input.

    For example, if \(f\) equals the line \(x + 3\) away from \(c\), then \(F\) is that line and needs no piecewise rule.

When it fails

If the left limit and the right limit are different, no value for \(F(c)\) equals both. The dot meets the left piece, and the red segment shows the gap that remains on the right.

If one side rises without bound, no real number equals its limit, so \(f\) has no continuous extension at \(c\).

Example 1: write the new function

Write the continuous extension of \(f(x) = \dfrac{2x^2 - x - 6}{x - 2}\) at \(x = 2\).

  1. Factor the top. For \(x \neq 2\), the factor \(x - 2\) cancels.

    \[\begin{aligned} f(x) &= \frac{(2x+3)(x-2)}{x-2} \\ &= 2x + 3 \end{aligned}\]

  2. Both sides approach \(2 \cdot 2 + 3 = 7\), so \(L = 7\).

    \[\lim_{x\to 2} f(x) = \cy{7}\]

  3. Give \(F\) the value 7 at \(x = 2\), and let it copy \(f\) everywhere else.

    \[F(x) = \begin{cases} f(x) & x \neq 2 \\ \cy{7} & x = 2 \end{cases}\]

  4. Because \(2 \cdot 2 + 3 = 7\), \(F(x) = 2x + 3\) at every \(x\), including 2, so the graph of \(F\) is a line with no break.

Example 2: a value is already there

Let \(k(x) = \dfrac{x^2 - 5x + 6}{x - 3}\) for \(x \neq 3\), and \(k(3) = 4\). Change only the value at \(x = 3\) so that \(k\) is continuous at 3.

  1. The top is \((x - 2)(x - 3)\), so \(k(x) = x - 2\) for \(x \neq 3\). Both sides approach \(3 - 2 = 1\).

    \[\lim_{x\to 3} k(x) = \cy{1}\]

  2. A tempting wrong move: The function already has a value at 3, so keep \(k(3) = 4\).

    \[k(3) = 4\]

    The dot at height 4 is 3 units above the ring. The red segment shows this gap, so the graph still has a break at 3.

  3. Replace the value at 3 with the limit 1. The dot moves into the ring, and the new \(k\) equals \(x - 2\) at every input.

    \[k(3) = \cy{1}\]

Practice

Let \(g(x) = \dfrac{4x^2 - 9}{2x + 3}\) for \(x \neq -\tfrac32\). Find the value \(G(-\tfrac32)\) that makes the extension \(G\) continuous.

Move the dot to the height that the two sides approach. Factor the top to find that height.

Solution

  1. The top is a difference of squares. For \(x \neq -\tfrac32\), the factor \(2x + 3\) cancels.

    \[\begin{aligned} g(x) &= \frac{(2x-3)(2x+3)}{2x+3} \\ &= 2x - 3 \end{aligned}\]

  2. Both sides approach \(2(-\tfrac32) - 3 = -6\), so the extension takes this value.

    \[G\!\left(-\tfrac32\right) = \cy{-6}\]

Questions

The graph shows \(f(x) = x^2 + 2\) with a ring on the curve at \(x = 1\). A separate dot sits above \(x = 1\) at height 5.

  1. Which value at \(x = 1\) makes the function continuous at 1?

    • 3
    • 5
    • 1
    • 2
    Answer

    3

  2. The extension puts a dot in the ring. Which part of the graph is different from before?

    • Only the point above \(x = 1\)
    • The whole curve near \(x = 1\)
    • Every point to the right of \(x = 1\)
    Answer

    Only the point above \(x = 1\)

  3. Let \(f(x) = \dfrac{x^2 - 4x - 21}{x - 7}\) for \(x \neq 7\), and let \(F\) be its continuous extension. Find \(F(7)\).

    • 10
    • 0
    • 7
    • \(-3\)
    Answer

    10

Summary

You find the limit at the missing point, give the new function that value there, and keep every other value of the old function.

Only one point changes, so the limit stays the same and equals the new value, which makes the function continuous there.

You build a continuous extension when a formula has one missing point and you need a function with no gap.

This is one of four skills in topic 1.13 of Which & Why: AP Calculus.