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You are watching: A function f is continuous on the closed interval

lot of limit analysis relates come a ide known as **continuity**. A duty is said to it is in **continuous** on one interval when the duty is identified at every suggest on the interval and undergoes no interruptions, jumps, or breaks. If some role f(x) satisfies this criteria from x=a to x=b, because that example, we say the f(x) is continuous on the interval . The brackets mean that the term is **closed** -- the it contains the endpoints a and also b. In other words, the the interval is identified as a ≤ x ≤ b. One **open** term (a, b), ~ above the other hand, would certainly not incorporate endpoints a and also b, and also would be defined as a Toggle Explanation Toggle line Numbers 1) Plot x2 native x=0 to x=1 2) Plot -x+2 from x=1 come x=2 3) Plot x2-3*x+2 from x=2 to x=3 4) produce a black suggest at (0, 0) 5) Create one more black allude at (3, 2) 6) integrate the plots and points right into a solitary graph with the given bounds

The role f(x) in the graph is known as a **piecewise** function, or one that has actually multiple, well, pieces. As you have the right to see, the role travels from x=0 to x=3 without interruption, and also since the 2 endpoints space closed (designated by the filled-in black circles), f(x) is continuous on the close up door interval <0, 3>. To think that it one more way, if you can trace a duty on one interval there is no picking up your pen (and without running over any holes), the role is consistent on that interval.

utilizing our expertise of boundaries from the vault lesson, we can say that:

and

The Intermediate value Theorem

among the much more important theorems relating to consistent functions is the **Intermediate value Theorem**, which claims that if a role f is consistent on a closeup of the door interval and k is any kind of number in between f(a) and f(b), climate there should exist at the very least one number c such the f(c) = k. In other words, if f is constant on , it need to pass with every y-value bounded by f(a) and also f(b). In the continuous function graphed above, for example, f(0) = 0 and also f(3) = 2, for this reason f(x) must pass with all y-values bounded by and including 0 and also 2 top top the term <0, 3>, which as one can see, the does.

Look in ~ this example, now, that a role that is *not* constant on the interval because that which that 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 indigenous x=0 to x=1 2) Plot -x+3 from x=1 come x=2 3) Plot -(x-3)3+2 indigenous x=2 to x=3 4) create a dashed line to show that the function jumps come a y-value of 3 when x is same to 1 5-10) develop open (faceted) or closed (filled-in) point out to indicate whether the intervals are open up or closeup of the door 11) integrate the plots, line, and points right into a graph v the given bounds

The role shown in the graph is *not* consistent on the closed interval <0, 3>, due to the fact that it has **discontinuities** at both x=1 and also x=2. A discontinuity is any x-value at which a role has an interruption, break or run -- something that would call for you to choose up your pen if you to be tracing the function. The filled-in black circles, again, indicate that the interval includes that point, if the open circles show that the term excludes that point. The dashed line at x=1 mirrors that f(1) = 3, not 2.

because the above graph contends least one discontinuity, that does not satisfy the demands of the Intermediate worth Theorem and also therefore *does not* have to pass v every y-value between f(0) and also f(3).

also though f(1) = 3, however, the border of f(x) as x approaches 1 does *not* same 3, since the role approaches a worth of 2 native both political parties of x=1. However what around the limit of f(x) as x philosophies 2? The duty does not strategy one particular value from either side of x=2, but we deserve to still explain its behavior from the best or the left. This is the basis of one-sided limits, which is the topic of the next section.

## Practice Problems

recognize which of the following functions are consistent on the close up door intervals because that which they space shown. For functions that space not continuous, identify the x-coordinates of your discontinuities. 1) # simply a note -- sin(x) revolutionized doesn"t in reality follow the very same curve as x2p1 = 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() Toggle answer Toggle line Numbers 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 price Toggle line Numbers 3) # exp() method "e come 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 price Toggle heat Numbers determine the value or values of k that will make every function constant on the offered interval.* 4) Toggle prize 5) Toggle answer 6) Toggle price use the Intermediate worth Theorem to demonstrate that a equipment exists to each equation ~ above the term given. 7) ;<1, 3> 0, so a number c need to exist in <1, 3> such the f(c) = 0");">Toggle price 8) ;<-1, 1> 2 and f(1) Toggle answer 9) ;<1, 2> 5 to find that f(1) 0, so a number c should exist in <1, 1> such that f(c) = 0");">Toggle price *As a thedesigningfairy.com-based method of checking her answers, you can use solve(...==..., k) to settle the provided equation because that k. For example, to resolve for x as soon as the expression x2-4*x+4 is equal to 0, you would certainly use:solve(x^2-4*x+4==0, x) Two equates to signs (as offered here) signifies a test for equality, if a single equals sign signifies assignment. For example, x==1 method "does x same 1?" vice versa, x=1 method "assign 1 to x".

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