what value must be chosen for a to make this function continuous

2. Limits

2.four Continuity

Learning Objectives

Explicate the 3 conditions for continuity at a point.
Describe three kinds of discontinuities.
Ascertain continuity on an interval.
State the theorem for limits of composite functions.
Provide an instance of the intermediate value theorem.

Many functions accept the belongings that their graphs can exist traced with a pencil without lifting the pencil from the folio. Such functions are called continuous. Other functions have points at which a pause in the graph occurs, merely satisfy this property over intervals contained in their domains. They are continuous on these intervals and are said to accept a discontinuity at a signal where a suspension occurs.

We begin our investigation of continuity by exploring what it ways for a function to take continuity at a indicate. Intuitively, a function is continuous at a particular bespeak if there is no suspension in its graph at that point.

Continuity at a Point

Before nosotros expect at a formal definition of what it means for a function to be continuous at a point, let's consider diverse functions that fail to meet our intuitive notion of what it means to be continuous at a point. We then create a list of conditions that prevent such failures.

Our first function of interest is shown in (Figure). We see that the graph of $f(x)$ has a hole at $a$ . In fact, $f(a)$ is undefined. At the very to the lowest degree, for $f(x)$ to be continuous at $a$ , we demand the following conditions:

i. $f(a)$ is defined.

However, as we see in (Figure), this status alone is insufficient to guarantee continuity at the point $a$ . Although $f(a)$ is defined, the function has a gap at $a$ . In this example, the gap exists because $\underset{x\to a}{\lim}f(x)$ does not exist. We must add another condition for continuity at $a$ —namely,

two. $\underset{x\to a}{\lim}f(x)$ exists.

However, as we see in (Effigy), these ii conditions by themselves exercise not guarantee continuity at a point. The function in this figure satisfies both of our starting time two conditions, but is however not continuous at $a$ . We must add together a 3rd condition to our list:

iii. $\underset{x\to a}{\lim}f(x)=f(a)$ .

Now nosotros put our listing of conditions together and form a definition of continuity at a bespeak.

The following procedure can be used to analyze the continuity of a function at a point using this definition.

The next iii examples demonstrate how to apply this definition to decide whether a office is continuous at a given indicate. These examples illustrate situations in which each of the weather condition for continuity in the definition succeed or fail.

Determining Continuity at a Indicate, Condition ane

Using the definition, determine whether the function $f(x)=(x^2-4)/(x-2)$ is continuous at $x=2$ . Justify the conclusion.

Determining Continuity at a Signal, Condition 2

Using the definition, determine whether the function $f(x)=\begin{cases} -x^2+4 & \text{if} \, x \le 3 \\ 4x-8 & \text{if} \, x > 3 \end{cases}$ is continuous at $x=3$ . Justify the conclusion.

Solution

Let'due south brainstorm by trying to summate $f(3)$ .

$f(3)=-(3^2)+4=-5$ .

Thus, $f(3)$ is defined. Next, we calculate $\underset{x\to 3}{\lim}f(x)$ . To do this, we must compute $\underset{x\to 3^-}{\lim}f(x)$ and $\underset{x\to 3^+}{\lim}f(x)$ :

$\underset{x\to 3^-}{\lim}f(x)=-(3^2)+4=-5$

and

$\underset{x\to 3^+}{\lim}f(x)=4(3)-8=4$ .

Therefore, $\underset{x\to 3}{\lim}f(x)$ does not be. Thus, $f(x)$ is not continuous at 3. The graph of $f(x)$ is shown in (Figure).

Determining Continuity at a Point, Status 3

Using the definition, make up one's mind whether the function $f(x)=\begin{cases} \frac{\sin x}{x} & \text{if} \, x \ne 0 \\ 1 & \text{if} \, x = 0 \end{cases}$ is continuous at $x=0$ .

Solution

First, detect that

$f(0)=1$ .

Next,

$\underset{x\to 0}{\lim}f(x)=\underset{x\to 0}{\lim}\frac{\sin x}{x}=1$ .

Last, compare $f(0)$ and $\underset{x\to 1}{\lim}f(x)$ . We encounter that

$f(0)=1=\underset{x\to 0}{\lim}f(x)$ .

Since all 3 of the atmospheric condition in the definition of continuity are satisfied, $f(x)$ is continuous at $x=0$ .

Using the definition, determine whether the office $f(x)=\begin{cases} 2x+1 & \text{if} \, x < 1 \\ 2 & \text{if} \, x = 1 \\ -x+4 & \text{if} \, x > 1 \end{cases}$ is continuous at $x=1$ . If the office is not continuous at 1, indicate the condition for continuity at a betoken that fails to hold.

Solution

$f$ is not continuous at ane because $f(1)=2\ne 3=\underset{x\to 1}{\lim}f(x)$ .

By applying the definition of continuity and previously established theorems concerning the evaluation of limits, we can land the following theorem.

Continuity of Polynomials and Rational Functions

Polynomials and rational functions are continuous at every indicate in their domains.

Types of Discontinuities

As nosotros have seen in (Figure) and (Figure), discontinuities have on several different appearances. We classify the types of discontinuities we take seen thus far as removable discontinuities, infinite discontinuities, or jump discontinuities. Intuitively, a removable discontinuity is a discontinuity for which there is a hole in the graph, a jump discontinuity is a noninfinite discontinuity for which the sections of the part practise not meet up, and an infinite aperture is a discontinuity located at a vertical asymptote. (Effigy) illustrates the differences in these types of discontinuities. Although these terms provide a handy way of describing three common types of discontinuities, keep in mind that non all discontinuities fit neatly into these categories.

0, there is an open circle on the line and a closed circle a few units above the line. The second is a jump discontinuity. Here, there are 2 lines with positive slope. The kickoff line exists for x<=a, and the second exists for ten>a, where a>0. The first line ends at a solid circle where x=a, and the second begins a few units up with an open up circle at 10=a. The third discontinuity type is infinite discontinuity. Here, the function has two parts separated by an asymptote 10=a. The first segment is a curve stretching along the x axis to 0 every bit 10 goes to negative infinity and forth the y axis to infinity as 10 goes to zero. The second segment is a curve stretching along the y axis to negative infinity every bit x goes to nothing and forth the 10 axis to 0 as ten goes to infinity." width="975″ acme="315″> Figure vi. Discontinuities are classified as (a) removable, (b) leap, or (c) space.

These three discontinuities are formally defined equally follows:

Classifying a Discontinuity

In (Figure), we showed that $f(x)=\frac{x^2-4}{x-2}$ is discontinuous at $x=2$ . Classify this discontinuity equally removable, jump, or infinite.

Solution

To allocate the discontinuity at 2 we must evaluate $\underset{x\to 2}{\lim}f(x)$ :

$\begin{array}{cc}\underset{x\to 2}{\lim}f(x) & =\underset{x\to 2}{\lim}\frac{x^2-4}{x-2} \\ & =\underset{x\to 2}{\lim}\frac{(x-2)(x+2)}{x-2} \\ & =\underset{x\to 2}{\lim}(x+2) \\ & = 4 \end{array}$

Since $f$ is discontinuous at ii and $\underset{x\to 2}{\lim}f(x)$ exists, $f$ has a removable discontinuity at $x=2$ .

Classifying a Discontinuity

In (Figure), we showed that $f(x)=\begin{cases} -x^2+4 & \text{if} \, x \le 3 \\ 4x-8 & \text{if} \, x > 3 \end{cases}$ is discontinuous at $x=3$ . Classify this discontinuity as removable, jump, or infinite.

Classifying a Aperture

Determine whether $f(x)=\frac{x+2}{x+1}$ is continuous at −i. If the function is discontinuous at −one, classify the aperture as removable, jump, or infinite.

For $f(x)=\begin{cases} x^2 & \text{if} \, x \ne 1 \\ 3 & \text{if} \, x = 1 \end{cases}$ , decide whether $f$ is continuous at 1. If $f$ is non continuous at 1, classify the discontinuity as removable, leap, or infinite.

Solution

Discontinuous at 1; removable

Continuity over an Interval

Now that we have explored the concept of continuity at a point, we extend that idea to continuity over an interval. Every bit we develop this idea for unlike types of intervals, information technology may be useful to keep in mind the intuitive idea that a function is continuous over an interval if we can utilize a pencil to trace the function between any two points in the interval without lifting the pencil from the paper. In training for defining continuity on an interval, nosotros brainstorm by looking at the definition of what it means for a function to be continuous from the right at a point and continuous from the left at a signal.

A function is continuous over an open interval if information technology is continuous at every betoken in the interval. A role $f(x)$ is continuous over a closed interval of the form $[a,b]$ if it is continuous at every betoken in $(a,b)$ and is continuous from the right at $a$ and is continuous from the left at $b$ . Analogously, a function $f(x)$ is continuous over an interval of the form $(a,b]$ if it is continuous over $(a,b)$ and is continuous from the left at $b$ . Continuity over other types of intervals are defined in a like fashion.

Requiring that $\underset{x\to a^+}{\lim}f(x)=f(a)$ and $\underset{x\to b^-}{\lim}f(x)=f(b)$ ensures that we can trace the graph of the role from the bespeak $(a,f(a))$ to the point $(b,f(b))$ without lifting the pencil. If, for case, $\underset{x\to a^+}{\lim}f(x)\ne f(a)$ , nosotros would need to lift our pencil to leap from $f(a)$ to the graph of the balance of the function over $(a,b]$ .

Continuity on an Interval

State the interval(s) over which the function $f(x)=\frac{x-1}{x^2+2x}$ is continuous.

Continuity over an Interval

State the interval(southward) over which the function $f(x)=\sqrt{4-x^2}$ is continuous.

State the interval(southward) over which the function $f(x)=\sqrt{x+3}$ is continuous.

Solution

$[-3,+\infty)$

The (Figure) allows u.s.a. to expand our ability to compute limits. In particular, this theorem ultimately allows us to demonstrate that trigonometric functions are continuous over their domains.

Composite Role Theorem

If $f(x)$ is continuous at $L$ and $\underset{x\to a}{\lim}g(x)=L$ , so

$\underset{x\to a}{\lim}f(g(x))=f(\underset{x\to a}{\lim}g(x))=f(L)$ .

Before we move on to (Figure), recollect that earlier, in the department on limit laws, we showed $\underset{x\to 0}{\lim} \cos x=1= \cos (0)$ . Consequently, we know that $f(x)= \cos x$ is continuous at 0. In (Figure) nosotros encounter how to combine this consequence with the composite role theorem.

Limit of a Composite Cosine Function

Evaluate $\underset{x\to \pi/2}{\lim}\cos(x-\frac{\pi }{2})$ .

Evaluate $\underset{x\to \pi}{\lim}\sin(x-\pi)$ .

The proof of the next theorem uses the composite function theorem likewise every bit the continuity of $f(x)= \sin x$ and $g(x)= \cos x$ at the signal 0 to show that trigonometric functions are continuous over their entire domains.

Continuity of Trigonometric Functions

Trigonometric functions are continuous over their entire domains.

Proof

Nosotros brainstorm past demonstrating that $\cos x$ is continuous at every real number. To do this, we must show that $\underset{x\to a}{\lim}\cos x = \cos a$ for all values of $a$ .

$\begin{array}{lllll}\underset{x\to a}{\lim}\cos x & =\underset{x\to a}{\lim}\cos((x-a)+a) & & & \text{rewrite} \, x \, \text{as} \, x-a+a \, \text{and group} \, (x-a) \\ & =\underset{x\to a}{\lim}(\cos(x-a)\cos a - \sin(x-a)\sin a) & & & \text{apply the identity for the cosine of the sum of two angles} \\ & = \cos(\underset{x\to a}{\lim}(x-a)) \cos a - \sin(\underset{x\to a}{\lim}(x-a))\sin a & & & \underset{x\to a}{\lim}(x-a)=0, \, \text{and} \, \sin x \, \text{and} \, \cos x \, \text{are continuous at 0} \\ & = \cos(0)\cos a - \sin(0)\sin a & & & \text{evaluate cos(0) and sin(0) and simplify} \\ & =1 \cdot \cos a - 0 \cdot \sin a = \cos a \end{array}$

The proof that $\sin x$ is continuous at every real number is analogous. Because the remaining trigonometric functions may exist expressed in terms of $\sin x$ and $\cos x,$ their continuity follows from the quotient limit police force. $_\blacksquare$

As you lot can meet, the composite role theorem is invaluable in demonstrating the continuity of trigonometric functions. As nosotros continue our study of calculus, we revisit this theorem many times.

The Intermediate Value Theorem

Functions that are continuous over intervals of the form $[a,b]$ , where $a$ and $b$ are real numbers, exhibit many useful properties. Throughout our report of calculus, we will encounter many powerful theorems apropos such functions. The showtime of these theorems is the Intermediate Value Theorem.

Application of the Intermediate Value Theorem

Show that $f(x)=x- \cos x$ has at least one zero.

When Can You Apply the Intermediate Value Theorem?

When Can You lot Apply the Intermediate Value Theorem?

Show that $f(x)=x^3-x^2-3x+1$ has a null over the interval $[0,1]$ .

Key Concepts

For the following exercises, determine the betoken(southward), if any, at which each function is discontinuous. Allocate whatever aperture as jump, removable, infinite, or other.

one. $f(x)=\frac{1}{\sqrt{x}}$

[reveal-respond q="955865″]Prove Solution[/reveal-answer]
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The role is defined for all $x$ in the interval $(0,\infty)$ .

2. $f(x)=\frac{2}{x^2+1}$

three. $f(x)=\frac{x}{x^2-x}$

Solution

Removable discontinuity at $x=0$ ; infinite discontinuity at $x=1$

4. $g(t)=t^{-1}+1$

five. $f(x)=\frac{5}{e^x-2}$

Solution

Infinite discontinuity at $x=\ln 2$

half-dozen. $f(x)=\frac{|x-2|}{x-2}$

7. $H(x)= \tan 2x$

Solution

Space discontinuities at $x=\frac{(2k+1)\pi}{4}$ , for $k=0, \, \pm 1, \, \pm 2, \, \pm 3, \cdots$

8. $f(t)=\frac{t+3}{t^2+5t+6}$

For the post-obit exercises, make up one's mind if the part continuous at the given point. If it is discontinuous, what type of aperture is information technology?

ix. $\frac{2x^2-5x+3}{x-1}$ at $x=1$

Solution

No. Information technology is a removable discontinuity.

ten. $h(\theta)=\frac{\sin \theta - \cos \theta}{\tan \theta}$ at $\theta =\pi$

11. $g(u)=\begin{cases} \frac{6u^2+u-2}{2u-1} & \text{if} \, u \ne \frac{1}{2} \\ \frac{7}{2} & \text{if} \, u = \frac{1}{2} \end{cases}$ at $u=\frac{1}{2}$

Solution

Aye. It is continuous.

12. $f(y)=\frac{\sin(\pi y)}{\tan(\pi y)}$ , at $y=1$

13. $f(x)=\brainstorm{cases} x^2-due east^x & \text{if} \, x < 0 \\ x-1 & \text{if} \, x \ge 0 \end{cases}$ at $x=0$

Solution

Yes. It is continuous.

14. $f(x)=\begin{cases} x \sin x & \text{if} \, x \le \pi \\ x \tan x & \text{if} \, x > \pi \end{cases}$ at $x=\pi$

In the following exercises, observe the value(s) of $k$ that makes each office continuous over the given interval.

15. $f(10)=\brainstorm{cases} 3x+ii & \text{if} \, x < k \\ 2x-3 & \text{if} \, k \le x \le 8 \end{cases}$

Solution

$k=-5$

16. $f(\theta)=\begin{cases} \sin \theta & \text{if} \, 0 \le \theta < \frac{\pi}{2} \\ \cos (\theta + k) & \text{if} \, \frac{\pi}{2} \le \theta \le \pi \end{cases}$

17. $f(x)=\begin{cases} \frac{x^2+3x+2}{x+2} & \text{if} \, x \ne -2 \\ k & \text{if} \, x = -2 \end{cases}$

Solution

$k=-1$

18. $f(x)=\brainstorm{cases} e^{kx} & \text{if} \, 0 \le x < 4 \\ x+3 & \text{if} \, 4 \le x \le 8 \end{cases}$

xix. $f(ten)=\begin{cases} \sqrt{kx} & \text{if} \, 0 \le x \le 3 \\ x+1 & \text{if} \, three < x \le 10 \end{cases}$

Solution

$k=\frac{16}{3}$

In the post-obit exercises, use the Intermediate Value Theorem (IVT).

22. [T] Use the statement "The cosine of $t$ is equal to $t$ cubed."

Write the statement equally a mathematical equation.
Prove that the equation in office a. has at least ane real solution.
Use a reckoner to find an interval of length 0.01 that contains a solution of the equation.

24.Consider the graph of the function $y=f(x)$ shown in the post-obit graph.

Observe all values for which the role is discontinuous.
For each value in part a., use the formal definition of continuity to explain why the function is discontinuous at that value.
Classify each discontinuity equally either jump, removable, or infinite.

Solution

1. It beings with an open circle at (1,3).">
b. It is not possible to redefine $f(1)$ since the discontinuity is a spring discontinuity.

Solution

Answers may vary; see the following example:

3.">

In the following exercises, suppose $y=f(x)$ is defined for all $x$ . For each description, sketch a graph with the indicated property.

29.Discontinuous at $x=1$ with $\underset{x\to -1}{\lim}f(x)=-1$ and $\underset{x\to 2}{\lim}f(x)=4$

Solution

Answers may vary; run into the following example:

1. It begins at (i,3).">

30.Discontinuous at $x=2$ but continuous elsewhere with $\underset{x\to 0}{\lim}f(x)=\frac{1}{2}$

Determine whether each of the given statements is truthful. Justify your response with an explanation or counterexample.

31. $f(t)=\frac{2}{e^t-e^{-t}}$ is continuous everywhere.

Solution

Faux. It is continuous over $(−\infty,0) \cup (0,\infty)$ .

33.If a function is not continuous at a point, so it is not defined at that bespeak.

Solution

False. Consider $f(x)=\begin{cases} x & \text{if} \, x \ne 0 \\ 4 & \text{if} \, x = 0 \end{cases}$

34.According to the IVT, $\cos x - \sin x - x = 2$ has a solution over the interval $[-1,1]$ .

36.The part $f(x)=\frac{x^2-4x+3}{x^2-1}$ is continuous over the interval $[0,3]$ .

The following problems consider the scalar grade of Coulomb's law, which describes the electrostatic force between 2 point charges, such as electrons. It is given past the equation $F(r)=k_e\frac{|q_1q_2|}{r^2}$ , where $k_e$ is Coulomb'due south abiding, $q_i$ are the magnitudes of the charges of the two particles, and $r$ is the distance betwixt the two particles.

Retrieve the discussion on spacecraft from the affiliate opener. The post-obit problems consider a rocket launch from Earth's surface. The force of gravity on the rocket is given by $F(d)=-mk/d^2$ , where $m$ is the mass of the rocket, $d$ is the altitude of the rocket from the center of Earth, and $k$ is a constant.

forty. [T] Determine the value and units of $k$ given that the mass of the rocket on World is 3 million kg. (Hint: The distance from the eye of Earth to its surface is 6378 km.)

41. [T] Later a certain distance $D$ has passed, the gravitational effect of Earth becomes quite negligible, and so nosotros tin approximate the forcefulness function by $F(d)=\brainstorm{cases} -\frac{mk}{d^2} & \text{if} \, d < D \\ 10,000 & \text{if} \, d \ge D \end{cases}$ Find the necessary condition $D$ such that the force office remains continuous.

Solution

$D=63.78$ km

Bear witness the following functions are continuous everywhere

43. $f(\theta) = \sin \theta$

44. $g(x)=|x|$

45.Where is $f(x)=\begin{cases} 0 & \text{if} \, x \, \text{is irrational} \\ 1 & \text{if} \, x \, \text{is rational} \end{cases}$ continuous?

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Source: https://opentextbc.ca/calculusv1openstax/chapter/continuity/

what value must be chosen for a to make this function continuous

2.four Continuity

Learning Objectives

Continuity at a Point

Determining Continuity at a Indicate, Condition ane

Determining Continuity at a Signal, Condition 2

Solution

Determining Continuity at a Point, Status 3

Solution

Solution

Continuity of Polynomials and Rational Functions

Types of Discontinuities

Classifying a Discontinuity

Solution

Classifying a Discontinuity

Classifying a Aperture

Solution

Continuity over an Interval

Continuity on an Interval

Continuity over an Interval

Solution

Composite Role Theorem

Limit of a Composite Cosine Function

Continuity of Trigonometric Functions

Proof

The Intermediate Value Theorem

Application of the Intermediate Value Theorem

When Can You Apply the Intermediate Value Theorem?

When Can You lot Apply the Intermediate Value Theorem?

Key Concepts

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

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