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<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />

Section 1.3 • Continuity 105<br />

2<br />

y<br />

20<br />

10<br />

(3, 9)<br />

2 4<br />

x 2 9<br />

if x 3<br />

x 3<br />

f (x) <br />

9 if x 3<br />

x 2 3 if x 3<br />

Figure 24 f is discontinuous at 3.<br />

IN WORDS Visually, the graph of a<br />

function f with a removable<br />

discontinuity has a hole at (c, lim x→c<br />

f (x)).<br />

The discontinuity is removable because<br />

we can fill in the hole by appropriately<br />

defining f . The function will then be<br />

continuous at x = c.<br />

NEED TO REVIEW? The floor function<br />

is discussed in Section P.2, p. 19.<br />

y<br />

2<br />

2<br />

2<br />

Figure 25 f (x) =x<br />

2<br />

4<br />

x<br />

x<br />

Solution Since f (3) = 9, the function f is defined at 3. To check the second condition,<br />

we investigate the one-sided limits.<br />

Teaching Tip<br />

x 2 − 9<br />

lim f (x) = lim<br />

x→3 −<br />

x→3 − x − 3 = lim (x − 3)(x + 3)<br />

= lim<br />

x→3 − x − 3 x→3−(x + 3) = 6<br />

↑<br />

Divide out x − 3<br />

lim f (x) = lim<br />

x→3 +<br />

x→3 +(x 2 − 3) = 9 − 3 = 6<br />

3<br />

Since lim f (x) = lim f (x), then lim f (x) exists. But, lim f (x) = 6 and f (3) = 9,<br />

x −25x<br />

x→3 −<br />

x→3 +<br />

x→3 x→3 fx ( ) =<br />

2<br />

so the third condition of continuity is not satisfied. The function f is discontinuous<br />

x −5x<br />

at 3. ■<br />

Figure 24 shows the graph of f .<br />

NOW WORK Problem 25 and AP® Practice Problems 5, 8, and 11.<br />

fx<br />

The discontinuity at c = 3 in Example 3 is called a removable discontinuity because<br />

we can redefine f at the number c to equal lim f (x) and make f continuous at c. So, in<br />

x→c fx<br />

Example 3, if f (3) is redefined to be 6, then f would be continuous at 3.<br />

DEFINITION Removable Discontinuity<br />

Let f be a function that is defined everywhere in an open interval containing c, except<br />

fx<br />

possibly at c. The number c is called a removable discontinuity of f if the function<br />

is discontinuous at c but lim f (x) exists. The discontinuity is removed by defining (or<br />

x→c<br />

redefining) the value of f at c to be lim f (x).<br />

x→c fx<br />

NOW WORK Problems 13 and 35 and AP® Practice Problem 4.<br />

Determining Whether a Function Is Continuous<br />

EXAMPLE 4 at a Number<br />

Determine whether the floor function f (x) =x is continuous at 1.<br />

Solution The floor function f (x) =x =the greatest integer ≤ x. The floor function f<br />

is defined at 1 and f (1) = 1. But<br />

lim f (x) = lim<br />

x→1 −<br />

x→1−x =0 and lim f (x) = lim<br />

x→1 +<br />

x→1 +x<br />

=1<br />

So, limx does not exist. Since limx does not exist, f is discontinuous at 1. ■<br />

x→1 x→1<br />

Figure 25 illustrates that the floor function is discontinuous at each integer. Also,<br />

none of the discontinuities of the floor function is removable. Since at each integer the<br />

value of the floor function “jumps” to the next integer, without taking on any intermediate<br />

values, the discontinuity at integer values is called a jump discontinuity.<br />

evaluate fc<br />

NOW WORK Problem 53.<br />

We have defined what it means for a function f to be continuous at a number. Now<br />

we define one-sided continuity at a number.<br />

Teaching Tip<br />

DEFINITION One-Sided Continuity at a Number<br />

Let f be a function defined on the interval (a, c]. Then f is continuous from the left<br />

at the number c if<br />

lim f (x) = f (c)<br />

x→c −<br />

Let f be a function defined on the interval [c, b). Then f is continuous from the right<br />

at the number c if<br />

lim f (x) = f (c)<br />

x→c +<br />

TRM Section 1.3: Worksheet 2<br />

This worksheet contains 5 piecewise functions.<br />

The students are asked to determine if the<br />

function is continuous at various values of each<br />

function.<br />

Continuity can also be taught using<br />

graphical and analytic methods. Students<br />

may be surprised to see that the function<br />

looks like a line.<br />

Notice the algebraic simplification:<br />

3<br />

x −25x<br />

( ) =<br />

2<br />

x −5x<br />

2<br />

xx ( −25)<br />

( ) =<br />

xx ( −5)<br />

( x− 5)( x+<br />

5)<br />

( ) =<br />

( x −5)<br />

( ) = x + 5<br />

The x’s and the x – 5’s were removed from<br />

the equation. The values, 0 and 5, were<br />

not in the domain of the original function.<br />

Since both were removed through algebraic<br />

simplification, they are both removable<br />

discontinuities. This function has no other<br />

points of discontinuity.<br />

Common Error<br />

Remind the students that when they<br />

(), they must substitute the<br />

value c into the original function, not the<br />

simplified version of the original function.<br />

If you present the left–right–center check<br />

for continuity at a point, consider having<br />

the students evaluate f() c first. If f()<br />

c<br />

is undefined, the function cannot be<br />

continuous at that point.<br />

Teaching Tip<br />

If a student is given a function and asked to<br />

identify all points of discontinuity, have the<br />

student begin by identifying the domain of<br />

the function. Points not in the domain of the<br />

function are points of discontinuity.<br />

AP® Exam Tip<br />

Although questions about continuity<br />

seldom appear directly on the multiplechoice<br />

portion of the exam, it is<br />

common to see continuity as part of a<br />

multiple-choice question. For example,<br />

a principle of continuity may appear as<br />

one of the answer choices. Also, in a<br />

problem about continuity, it is common<br />

to be given a piecewise function.<br />

Section 1.3 • Continuity 105<br />

<strong>TE</strong>_<strong>Sullivan</strong>_Chapter01_PART I.indd 2<br />

11/01/17 9:57 am

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