Much of limit analysis relates to a concept known as **continuity**.
A function is said to be **continuous** on an interval when the
function is defined at every point on that interval and undergoes no
interruptions, jumps, or breaks. If some function f(x) satisfies these criteria
from x=a to x=b, for example, we say that f(x) is continuous on the interval
[a, b]. The brackets mean that the interval is **closed** -- that
it includes the endpoints a and b. In other words, that the interval is defined as
a ≤ x ≤ b. An **open** interval (a, b), on the other hand, would
not include endpoints a and b, and would be defined as a < x < b.

For an example of continuity, start a new worksheet called 02-Continuity, then recreate the following graph using the provided code. (The f(x)=... is superimposed).

p1 = plot(x^2, x, 0, 1) p2 = plot(-x+2, x, 1, 2) p3 = plot(x^2-3*x+2, x, 2, 3) pt1 = point((0, 0), rgbcolor='black', pointsize=30) pt2 = point((3, 2), rgbcolor='black', pointsize=30) (p1+p2+p3+pt1+pt2).show(xmin=0, xmax=3, ymin=0, ymax=2)Toggle Explanation Toggle Line Numbers

1) Plot x^{2} from x=0 to x=1

2) Plot -x+2 from x=1 to x=2

3) Plot x^{2}-3*x+2 from x=2 to x=3

4) Create a black point at (0, 0)

5) Create another black point at (3, 2)

6) Combine the plots and points into a single graph with the given bounds

The function f(x) in the graph is known as a **piecewise** function,
or one that has multiple, well, pieces. As you can see, the function travels
from x=0 to x=3 without interruption, and since the two endpoints are closed
(designated by the filled-in black circles), f(x) is continuous on the closed
interval [0, 3]. To think of it another way, if you can trace a function on an
interval without picking up your pen (and without running over any holes), the
function is continuous on that interval.

Using our knowledge of limits from the previous lesson, we can say that:

and

One of the more important theorems relating to continuous functions is
the **Intermediate Value Theorem**, which states that if a function
f is continuous on a closed interval [a, b] and k is any number between f(a) and
f(b), then there must exist at least one number c such that f(c) = k. In other
words, if f is continuous on [a, b], it must pass through every y-value bounded
by f(a) and f(b). In the continuous function graphed above, for example,
f(0) = 0 and f(3) = 2, so f(x) must pass through all y-values bounded by and
including 0 and 2 on the interval [0, 3], which as one can see, it does.

Look at this example, now, of a function that is *not* continuous on the
interval for which it is shown.

p1 = plot(2*x, x, 0, 1) p2 = plot(-x+3, x, 1, 2) p3 = plot(-(x-3)^3+2, x, 2, 3) l1 = line([(1, 2.1), (1, 2.9)], linestyle='--') pt1 = point((0, 0), rgbcolor='black', pointsize=30) pt2 = point((1, 2), rgbcolor='white', faceted=True, pointsize=30) pt3 = point((1, 3), rgbcolor='black', pointsize=30) pt4 = point((2, 1), rgbcolor='black', pointsize=30) pt5 = point((2, 3), rgbcolor='white', faceted=True, pointsize=30) pt6 = point((3, 2), rgbcolor='black', pointsize=30) (p1+p2+p3+l1+pt1+pt2+pt3+pt4+pt5+pt6).show(xmin=0, xmax=3, ymin=0, ymax=3)Toggle Explanation Toggle Line Numbers

1) Plot 2*x from x=0 to x=1

2) Plot -x+3 from x=1 to x=2

3) Plot -(x-3)^{3}+2 from x=2 to x=3

4) Create a dashed line to indicate that the function jumps to a y-value of 3
when x is equal to 1

5-10) Create open (faceted) or closed (filled-in) points to indicate whether the intervals are open
or closed

11) Combine the plots, line, and points into a graph with the given bounds

The function shown in the graph is *not* continuous on the closed interval
[0, 3], since it has **discontinuities** at both x=1 and x=2. A
discontinuity is any x-value at which a function has an interruption, break or
jump -- something that would require you to pick up your pen if you were tracing
the function. The filled-in black circles, again, indicate that
the interval includes that point, while the open circles indicate that the
interval excludes that point. The dashed line at x=1 shows that f(1) = 3, not 2.

Since the above graph has at least one discontinuity, it does not satisfy the
requirements of the Intermediate Value Theorem and therefore *does not*
have to pass through every y-value between f(0) and f(3).

Even though f(1) = 3, however, the limit of f(x) as x approaches 1 does *not*
equal 3, since the function approaches a value of 2 from both sides of x=1. But
what about the limit of f(x) as x approaches 2? The function does not approach
one specific value from either side of x=2, but we can still describe its behavior
from the right or the left. This is the basis of one-sided limits, which is the
topic of the next section.

Determine which of the following functions are continuous on the closed intervals
for which they are shown. For functions that are not continuous, determine the
x-coordinates of their discontinuities.

1)

2)

3)

Determine the value or values of k that will make each function continuous on the given interval.*

4)

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5)

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6)

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Use the Intermediate Value Theorem to demonstrate that a solution exists to each equation on the interval given.

7) ; [1, 3]

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8) ; [-1, 1]

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9) ; [1, 2]

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*As a SageMath-based way of checking your answers, you can use solve(...==..., k) to solve the given equation for k. For example, to solve for x when the expression x^{2}-4*x+4 is equal to 0, you would use:

1)

# Just a note -- sin(x) transformed doesn't actually follow the same curve as xToggle answer Toggle Line Numbers^{2}p1 = plot((x+1)^2-1, x, -2, 0) p2 = plot(5/4*sin(pi*x/2), x, 0, 2) p3 = plot((x-3)^2-1, x, 2, 4) pt1 = point((-2, 0), rgbcolor='black', pointsize=30) pt2 = point((4, 0), rgbcolor='black', pointsize=30) (p1+p2+p3+pt1+pt2).show()

2)

p = plot(tan(x), x, -pi, pi, randomize=False, plot_points=101) pt1 = point((-pi, 0), rgbcolor='black', pointsize=30) pt2 = point((pi, 0), rgbcolor='black', pointsize=30) (p+pt1+pt2).show(xmin=-pi, xmax=pi, ymin=-10, ymax=10)Toggle answer Toggle Line Numbers

3)

# exp() means 'e to the ___' p1 = plot(exp(x/5)-1, x, 0, e) p2 = plot(ln(x), x, e, e^2) pt1 = point((0, 0), rgbcolor='black', pointsize=30) pt2 = point((e, exp(e/5)-1), rgbcolor='black', pointsize=30) pt3 = point((e, 1), faceted=True, rgbcolor='white', pointsize=30) pt4 = point((e^2, 2), rgbcolor='black', pointsize=30) (p1+p2+pt1+pt2+pt3+pt4).show(ymin=0, ymax=2)Toggle answer Toggle Line Numbers

Determine the value or values of k that will make each function continuous on the given interval.*

4)

Toggle answer

5)

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6)

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Use the Intermediate Value Theorem to demonstrate that a solution exists to each equation on the interval given.

7) ; [1, 3]

Toggle answer

8) ; [-1, 1]

Toggle answer

9) ; [1, 2]

Toggle answer

*As a SageMath-based way of checking your answers, you can use solve(...==..., k) to solve the given equation for k. For example, to solve for x when the expression x

solve(x^2-4*x+4==0, x)Two equals signs (as used here) signifies a test for equality, while a single equals sign signifies assignment. For example, x==1 means 'does x equal 1?' whereas x=1 means 'assign 1 to x'.