Calculus Theorem About Continuity on Interval With Positive Constant R

\(\renewcommand{\mybuildpath}{./} \renewcommand{\labelitemi}{\scriptsize$\blacktriangleright$} \def\ds{\displaystyle} \def\R{\mathbb{R}} \def\arraystretch{2.5} \renewcommand{\Heq}{\overset{H}{=}} \renewcommand{\vect}{\textbf} \renewcommand{\longvect}{\overrightarrow} \newcommand{\diff}[2]{\dfrac{d#1}{d#2}} \newcommand{\diffp}[2]{\dfrac{\partial#1}{\partial#2}} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \)

¶

Subsection 3.7.1 Continuity

¶

The graph shown in Figure 3.3(a) represents a continuous function. Geometrically, this is because there are no jumps in the graphs. That is, if you pick a point on the graph and approach it from the left and right, the values of the function approach the value of the function at that point. For example, we can see that this is not true for function values near \(x=1\) on the graph in Figure 3.3(b) which is not continuous at that location.

Figure 3.3. (a) A continuous function. (b) A function with a discontinuity at \(x=1\text{.}\)

Definition 3.54. Continuous at a Point.

A function \(f\) is continuous at a point \(a\) if

\begin{equation*} \ds \lim_{x\to a} f(x) = f(a)\text{.} \end{equation*}

Some readers may prefer to think of continuity at a point as a three part definition. We provide the following guideline for determining the continuity of a function.

Guideline for Checking Continuity at a Point.

A function \(f(x)\) is continuous at \(x=a\) if the following three conditions hold:

1. \(f(a)\) is defined (that is, \(a\) belongs to the domain of \(f\)),

2. \(\ds{\lim_{x\to a}f(x)}\) exists (that is, left-hand limit \(=\) right-hand limit),

3. \(\ds{\lim_{x\to a}f(x)=f(a)}\) (that is, the numbers from 1 and 3 are equal).

The figures below show graphical examples of functions where either 1, 2 or 3 can fail to hold.

On the other hand, if \(f\) is defined on an open interval containing \(a\text{,}\) except perhaps at \(a\text{,}\) we can say that \(f\) is discontinuous at \(a\) if \(f\) is not continuous at \(a\text{.}\)

Graphically, you can think of continuity as being able to draw your function without having to lift your pencil off the paper. If your pencil has to jump off the page to continue drawing the function, then the function is not continuous at that point. This is illustrated in Figure 3.3(b) where if we tried to draw the function (from left to right) we need to lift our pencil off the page once we reach the point \(x=1\) in order to be able to continue drawing the function.

Definition 3.55. Continuity on an Open Interval.

A function \(f\) is continuous on an open interval \((a,b)\) if it is continuous at every point in the interval.

Furthermore, a function is everywhere continuous if it is continuous on the entire real number line \((-\infty,\infty)\text{..}\)

Recall the function graphed in a previous section as shown in Figure 3.4.

Figure 3.4. A function with discontinuities at \(x=-5\text{,}\) \(x=-2\text{,}\) \(x=-1\) and \(x=4\text{.}\)

We can draw this function without lifting our pencil except at the points \(x=-5\text{,}\) \(x=-2\text{,}\) \(x=-1\text{,}\) and \(x=4\text{.}\) Thus, \(f(x)\) is continuous at every real number except at these four numbers. At \(x=-5\text{,}\) \(x=-2\text{,}\) \(x=-1\text{,}\) and \(x=4\text{,}\) the function \(f(x)\) is discontinuous.

At \(x=-2\) we have a removable discontinuity because we could remove this discontinuity simply by redefining \(f(-2)\) to be \(3.5\text{.}\) Formally, we say \(f(x)\) has a removable discontinuity at \(x=a\) if \(\lim_{x\to a}f(x)\) exists but is not equal to \(f(a)\text{.}\) Note that we do not require \(f(a)\) to be defined in this case, that is, \(a\) need not belong to the domain of \(f(x)\text{.}\) At \(x=-5\) and \(x=-1\) we have jump discontinuities because the function jumps from one value to another. From the right of \(x=4\text{,}\) we have an infinite discontinuity because the function goes off to infinity. These types of discontinuities are summarized below.

Summary of Discontinuities.

Example 3.56. Continuous at a Point.

What value of \(c\) will make the following function \(f(x)\) continuous at \(2\text{?}\)

\begin{equation*} f(x)=\left\{\begin{array}{cl} \ds{\frac{x^2-x-2}{x-2}}\amp \qquad\mbox{if \(x\neq 2\)} \\ \\ c\amp \qquad\mbox{if \(x=2\)} \end{array} \right. \end{equation*}

Solution

In order to be continuous at \(2\) we require

\begin{equation*} \lim_{x\to 2}f(x)=f(2) \end{equation*}

to hold. We use the three part definition listed previously to check this.

1. First, \(f(2)=c\text{,}\) and \(c\) is some real number. Thus, \(f(2)\) is defined.

2. Now, we must evaluate the limit. Rather than computing both one-sided limits, we just compute the limit directly. For \(x\) close to \(2\) (but not equal to \(2\)) we can replace \(f(x)\) with \(\frac{x^2-x-2}{x-2}\) to get:

   \begin{equation*} \lim_{x\to 2}f(x)=\lim_{x\to 2}\frac{x^2-x-2}{x-2}=\lim_{x\to 2}\frac{(x-2)(x+1)}{x-2}=\lim_{x\to 2}(x+1)=3\text{.} \end{equation*}

   Therefore the limit exists and equals \(3\text{.}\)

3. Finally, for \(f\) to be continuous at \(2\text{,}\) we need that the numbers in the first two items to be equal. Therefore, we require \(c=3\text{.}\)

Thus, when \(c=3\text{,}\) \(f(x)\) is continuous at \(2\text{,}\) for any other value of \(c\text{,}\) \(f(x)\) is discontinuous at \(2\text{.}\)

For continuity on a closed interval, we consider the one-sided limits of a function. Recall that \(x\to a^-\) means \(x\) approaches \(a\) from values less than \(a\text{.}\)

Definition 3.57. Continuous from the Right and from the Left.

A function \(f\) is left continuous at a point \(a\) if

\begin{equation*} \ds \lim_{x\to a^-} f(x) = f(a) \end{equation*}

and right continuous at a point \(a\) if

\begin{equation*} \ds \lim_{x\to a^+} f(x) = f(a)\text{.} \end{equation*}

If a function \(f\) is continuous at \(a\text{,}\) then it is both left and right continuous at \(a\text{.}\)

The above definition regarding left (or right) continuous functions is illustrated with the following figure:

One-sided limits allows us to extend the definition of continuity to closed intervals. The following definition means a function is continuous on a closed interval if it is continuous in the interior of the interval and possesses the appropriate one-sided continuity at the endpoints of the interval.

Definition 3.58. Continuity on a Closed Interval.

A function \(f\) is continuous on the closed interval \([a,b]\) if:

1. it is continuous on the open interval \((a,b)\text{;}\)

2. it is left continuous at point \(a\text{:}\)

   \begin{equation*} \ds\lim_{x\to a^-}f(x)=f(a); \end{equation*}

   and

3. it is right continuous at point \(b\text{:}\)

   \begin{equation*} \ds\lim_{x\to b^+}f(x)=f(b)\text{.} \end{equation*}

This definition can be extended to continuity on half-open intervals such as \((a,b]\) and \([a,b)\text{,}\) and unbounded intervals.

Example 3.59. Continuity on Other Intervals.

The function \(f(x)=\sqrt{x}\) is continuous on the (closed) interval \([0,\infty)\text{.}\)

The function \(f(x)=\sqrt{4-x}\) is continuous on the (closed) interval \((-\infty,4]\text{.}\)

The continuity of functions is preserved under the operations of addition, subtraction, multiplication and division (in the case that the function in the denominator is nonzero).

Theorem 3.60. Operations of Continuous Functions.

If \(f\) and \(g\) are continous at \(a\text{,}\) and \(c\) is a constant, then the following functions are also continuous at \(a\text{:}\)

1. \(f\pm g\text{;}\)

2. \(cf\text{;}\)

3. \(fg\text{;}\)

4. \(f/g\) (provided \(g(a)\neq 0\)).

Below we list some common functions that are known to be continuous on every interval inside their domains.

Example 3.61. Common Types of Continuous Functions.

• Polynomials (for all \(x\)), e.g., \(y=mx+b\text{,}\) \(y=ax^2+bx+c\text{.}\)

• Rational functions (except at points \(x\) which gives division by zero).

• Root functions \(\sqrt[n]{x}\) (for all \(x\) if \(n\) is odd, and for \(x\geq 0\) if \(n\) is even).

• Trigonometric functions

• Inverse trigonometric functions

• Exponential funtions

• Logarithmic functions

For rational functions with removable discontinuities as a result of a zero, we can define a new function filling in these gaps to create a piecewise function that is continuous everywhere.

Continuous functions are where the direct substitution property hold. This fact can often be used to compute the limit of a continuous function.

Example 3.62. Evaluate a Limit.

Evaluate the following limit: \(\ds\lim_{x\to\pi}\frac{\sqrt{x}+\sin x}{1+x+\cos x}\text{.}\)

Solution

We will use a continuity argument to justify that direct substitution can be applied. By the list above, \(\sqrt{x}\text{,}\) \(\sin x\text{,}\) \(1\text{,}\) \(x\) and \(\cos x\) are all continuous functions at \(\pi\text{.}\) Then \(\sqrt x+\sin x\) and \(1+x+\cos x\) are both continuous at \(\pi\text{.}\) Finally,

\begin{equation*} \frac{\sqrt{x}+\sin x}{1+x+\cos x} \end{equation*}

is a continuous function at \(\pi\) since \(1+\pi+\cos\pi\neq 0\text{.}\) Hence, we can directly substitute to get the limit:

\begin{equation*} \lim_{x\to\pi}\frac{\sqrt{x}+\sin x}{1+x+\cos x}=\frac{\sqrt{\pi}+\sin \pi}{1+\pi+\cos \pi}=\frac{\sqrt{\pi}}{\pi}=\frac{1}{\sqrt{\pi}}\text{.} \end{equation*}

Continuity is also preserved under the composition of functions.

Theorem 3.63. Continuity of Function Composition.

If \(g\) is continuous at \(a\) and \(f\) is continuous at \(g(a)\text{,}\) then the composition function \(f\circ g\) is continuous at \(a\text{.}\)

Example 3.64. Continuity with Composition of Functions.

Determine where the following functions is continuous:

1. \(h(x)=\cos(x^2)\)

2. \(H(x)=\ln(1+\sin(x))\)

Solution

1. The functions that make up the composition \(h(x)=f(g(x))\) are \(g(x)=x^2\) and \(f(x)=\cos(x)\text{.}\) The function \(g\) is continuous on \(\mathbb{R}\) since it is a polynomial, and \(f\) is also continuous everywhere. Therefore, \(h(x)=(f\circ g)(x)\) is continuous on \(\mathbb{R}\) by Theorem 3.63.

2. We know from Example 3.61 that \(f(x)=\ln x\) is continuous and \(g(x)=1+\sin x\) are continuous. Thus by Theorem 3.63, \(H(x)=f(g(x))\) is continuous wherever it is defined. Now \(\ln(1+\sin x)\) is defined when \(1+\sin x>0\text{.}\) Recall that \(-1\leq \sin x\leq 1\text{,}\) so \(1+\sin x>0\) except when \(\sin x=-1\text{,}\) which happens when \(x=\pm 3\pi/2,\pm 7\pi/2,\ldots\text{.}\) Therefore, \(H\) has discontinuities when \(x=3\pi n/2\text{,}\) \(n=1,2,3,\ldots\) and is continuous on the intervals between these values.

Subsection 3.7.2 The Intermediate Value Theorem

¶

Whether or not an equation has a solution is an important question in mathematics. Consider the following two questions:

Example 3.65. Motivation for the Intermediate Value Theorem.

1. Does \(e^x+x^2=0\) have a solution?

2. Does \(e^x+x=0\) have a solution?

Solution

1. The first question is easy to answer since for any exponential function we know that \(a^x>0\text{,}\) and we also know that whenever you square a number you get a nonnegative answer: \(x^2\geq 0\text{.}\) Hence, \(e^x+x^2>0\text{,}\) and thus, is never equal to zero. Therefore, the first equation has no solution.

2. For the second question, it is difficult to see if \(e^x+x=0\) has a solution. If we tried to solve for \(x\text{,}\) we would run into problems. Let's make a table of values to see what kind of values we get (recall that \(e\approx 2.7183\)):

   \begin{equation*} \begin{array}{l|cccc} x \amp -2 \amp -1 \amp 0 \amp 1 \\ \hline e^x+x \amp e^{-2}-2 \approx -1.9 \amp e^{-1}-1 \approx 1 \amp e^0+0 = 1 \amp e + 1 \approx 3.7 \end{array} \end{equation*}

   Sketching this gives:

   

   Let \(f(x)=e^x+x\text{.}\) Notice that if we choose \(a=-1\) and \(b=0\) then we have \(f(a)\lt 0\) and \(f(b)>0\text{.}\) A point where the function \(f(x)\) crosses the \(x\)-axis gives a solution to \(e^x+x=0\text{.}\) Since \(f(x)=e^x+x\) is continuous (both \(e^x\) and \(x\) are continuous), then the function must cross the \(x\)-axis somewhere between \(-1\) and \(0\text{:}\)

   

   Therefore, our equation has a solution. Note that by looking at smaller and smaller intervals \((a,b)\) with \(f(a)\lt 0\) and \(f(b)>0\text{,}\) we can get a better and better approximation for a solution to \(e^x+x=0\text{.}\) For example, taking the interval \((-0.4,-0.6)\) gives \(f(-0.4)\lt 0\) and \(f(-0.6)>0\text{,}\) thus, there is a solution to \(f(x)=0\) between \(-0.4\) and \(-0.6\text{.}\) It turns out that the solution to \(e^x+x=0\) is \(x\approx -0.56714\text{.}\)

We now generalize the argument used in the previous example. In that example we had a continuous function that went from negative to positive and hence, had to cross the \(x\)-axis at some point. In fact, we don't need to use the \(x\)-axis, any line \(y=N\) will work so long as the function is continuous and below the line \(y=N\) at some point and above the line \(y=N\) at another point. This is known as the Intermediate Values Theorem and it is formally stated as follows:

Theorem 3.66. Intermediate Value Theorem.

If \(f\) is continuous on the interval \([a,b]\) and \(N\) is between \(f(a)\) and \(f(b)\text{,}\) where \(f(a)\neq f(b)\text{,}\) then there is a number \(c\) in \((a,b)\) such that \(f(c)=N\text{.}\)

The Intermediate Value Theorem guarantees that if \(f(x)\) is continuous and \(f(a)\lt N\lt f(b)\text{,}\) the line \(y=N\) intersects the function at some point \(x=c\text{.}\) Such a number \(c\) is between \(a\) and \(b\) and has the property that \(f(c)=N\) (see Figure 3.5(a)). We can also think of the theorem as saying if we draw the line \(y=N\) between the lines \(y=f(a)\) and \(y=f(b)\text{,}\) then the function cannot jump over the line \(y=N\text{.}\) On the other hand, if \(f(x)\) is not continuous, then the theorem may not hold. See Figure 3.5(b) where there is no number \(c\) in \((a,b)\) such that \(f(c)=N\text{.}\) Finally, we remark that there may be multiple choices for \(c\) (i.e., lots of numbers between \(a\) and \(b\) with \(y\)-coordinate \(N\)). See Figure 3.5(c) for such an example.

Figure 3.5. (a) A continuous function where IVT holds for a single value \(c\text{.}\) (b) A discontinuous function where IVT fails to hold. (c) A continuous function where IVT holds for multiple values in \((a,b)\text{.}\)

The Intermediate Value Theorem is most frequently used for \(N=0\text{.}\)

Example 3.67. Intermediate Value Theorem.

Show that there is a solution of \(\sqrt[3]{x}+x=1\) in the interval \((0,8)\text{.}\)

Solution

Let \(f(x)=\sqrt[3]{x}+x-1\text{,}\) \(N=0\text{,}\) \(a=0\text{,}\) and \(b=8\text{.}\) Since \(\sqrt[3]{x}\text{,}\) \(x\) and \(-1\) are continuous on \(\mathbb{R}\text{,}\) and the sum of continuous functions is again continuous, we have that \(f(x)\) is continuous on \(\mathbb{R}\text{,}\) thus in particular, \(f(x)\) is continuous on \([0,8]\text{.}\) We have \(f(a)=f(0)=\sqrt[3]{0}+0-1=-1\) and \(f(b)=f(8)=\sqrt[3]{8}+8-1=9\text{.}\) Thus \(N=0\) lies between \(f(a)=-1\) and \(f(b)=9\text{,}\) so the conditions of the Intermediate Value Theorem are satisfied. So, there exists a number \(c\) in \((0,8)\) such that \(f(c)=0\text{.}\) This means that \(c\) satisfies \(\sqrt[3]{c}+c-1=0\text{,}\) in otherwords, is a solution for the equation given.

Alternatively we can let \(f(x)=\sqrt[3]{x}+x\text{,}\) \(N=1\text{,}\) \(a=0\) and \(b=8\text{.}\) Then as before \(f(x)\) is the sum of two continuous functions, so is also continuous everywhere, in particular, continuous on the interval \([0,8]\text{.}\) We have \(f(a)=f(0)=\sqrt[3]{0}+0=0\) and \(f(b)=f(8)=\sqrt[3]{8}+8=10\text{.}\) Thus \(N=1\) lies between \(f(a)=0\) and \(f(b)=10\text{,}\) so the conditions of the Intermediate Value Theorem are satisfied. So, there exists a number \(c\) in \((0,8)\) such that \(f(c)=1\text{.}\) This means that \(c\) satisfies \(\sqrt[3]{c}+c=1\text{,}\) in otherwords, is a solution for the equation given.

Example 3.68. Roots of Function.

Explain why the function \(\ds f=x^3 + 3x^2+x-2\) has a root between 0 and 1.

Solution

By Theorem 3.9, \(f\) is continuous. Since \(f(0)=-2\) and \(f(1)=3\text{,}\) and \(0\) is between \(-2\) and \(3\text{,}\) there is a \(c\in(0,1)\) such that \(f(c)=0\text{.}\)

This example also points the way to a simple method for approximating roots.

Example 3.69. Approximating Roots.

Approximate the root of the previous example to one decimal place.

Solution

If we compute \(f(0.1)\text{,}\) \(f(0.2)\text{,}\) and so on, we find that \(f(0.6)\lt 0\) and \(f(0.7)>0\text{,}\) so by the Intermediate Value Theorem, \(f\) has a root between \(0.6\) and \(0.7\text{.}\) Repeating the process with \(f(0.61)\text{,}\) \(f(0.62)\text{,}\) and so on, we find that \(f(0.61)\lt 0\) and \(f(0.62)>0\text{,}\) so \(f\) has a root between \(0.61\) and \(0.62\text{,}\) and the root is \(0.6\) rounded to one decimal place.

Exercises for Section 3.7.

Exercise 3.7.1.

Determine the values of \(x\text{,}\) if any, at which each function is discontinuous. At each point of discontinuity, state the condition(s) for continuity which are violated.

1. \(f(x) = \begin{cases} x - 2 \amp x \leq 0 \\ 2 \amp x > 0 \end{cases}\)

   

   Answer

   \(x = 0\text{.}\) \(\lim\limits_{x \to 0}\) does not exist and is not equal to \(f(0)\text{.}\)

   Solution

   Away from \(x = 0\text{,}\) we see that \(f\) is continuous. Therefore, we look at \(x = 0\text{:}\)

   \begin{equation*} \begin{split} \amp \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} 2 = 2 \amp \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} x-2 = -2 \end{split} \end{equation*}

   Therefore, \(\lim\limits_{x \to 0}\) does not exist. We conclude that \(f\) is discontinuous at \(x = 0\text{.}\) In the graph of \(f\) which follows, this discontinuity is seen as a jump at \(x = 0\text{.}\)

2. \(f(x) = \begin{cases} x + 2 \amp x > 0 \\ -\frac{1}{2}x^{2}+2 \amp x \leq 0 \end{cases}\)

   

   Answer

   No points of discontinuity. \(f\) is continuous everywhere.

   Solution

   We again check continuity at \(x = 0\text{.}\)

   \begin{equation*} \begin{split} \amp \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} x + 2 = 2 \amp \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} \frac{1}{2}x^{2}+2 = 2 \end{split} \end{equation*}

   Therefore, \(\lim\limits_{x \to 0} = 2 = f(0) \implies f\) continuous at \(0\text{.}\) We again note that away from \(0\text{,}\) \(f\) is clearly continuous. There are therefore no points of discontinuity.

3. \(f(x) = \begin{cases} x + 2 \amp x > 0 \\ -5 \amp x = 0 \\ -\frac{1}{2}x^{2}+2 \amp x > 0 \end{cases}\)

   

   Answer

   \(x = 0\text{.}\) \(\lim\limits_{x \to 0} = 2\text{,}\) which is not equal to \(f(0) = 1\text{.}\)

   Solution

   We follow a simular procedure as above to find that \(\lim\limits_{x \to 0} f(x) = 2\text{.}\) However, \(f(0) = -5\text{,}\) and \(f\) is thus discontinuous at \(x = 0\text{.}\) In the graph of \(f\text{,}\) we see a 'hole' at \(x = 0\text{.}\)

4. \(f(x) = \begin{cases} \ln(x+2) \amp x \leq 1, x \neq -1 \\ -1 \amp x = -1 \\ -1 \amp x > 1 \end{cases}\)

   

   Answer

   \(x=\pm 1\text{.}\) \(\lim\limits_{x\to 1}\) DNE, \(\lim\limits_{x\to -1} \neq f(-1)\text{.}\)

   Solution

   At \(x=-1\text{:}\)

   \begin{equation*} \lim\limits_{x\to-1^+}f(x) = \lim\limits_{x\to-1^-}f(x) = 0\text{,} \end{equation*}

   thus \(\lim\limits_{x\to-1} f(x) = 0\text{.}\) However, \(f(-1) = -1 \neq 0\text{,}\) and so \(f\) is discontinuous at \(x=-1\text{.}\) At \(x=1\text{:}\)

   \begin{equation*} \lim\limits_{x\to 1^+}f(x) =-1 \neq \lim\limits_{x\to 1^-}f(x) = 1\text{,} \end{equation*}

   and so \(\lim\limits_{x\to 1^-}f(x)\) DNE. Thus, \(f\) is discontinuous at \(x=1\text{.}\) For all \(x \neq \pm 1\text{,}\) \(f\) is continuous.

Exercise 3.7.2.

Determine the values of \(x\) for which each function is continuous.

1. \(f(s) = \dfrac{2}{s^{2} + 1}\)

   Answer

   \(\left(-\infty,\infty\right)\)

   Solution

   Since \(s^2+1=0\) has no real solutions, we see that \(f\) is continuous for all \(x \in (-\infty,\infty)\text{.}\)

2. \(g(t) = \dfrac{2t+1}{t^{2}+t-2}\)

   Answer

   \(\left(-\infty, -2 \right) \cup \left(-2,1\right) \cup \left(1,\infty\right)\)

   Solution

   We notice that

   \begin{equation*} g(t) = \dfrac{2t+1}{t^2+t-2} = \dfrac{2t+1}{(t-1)(t+2)}\text{.} \end{equation*}

   Thus, \(g(t)\) is continuous for all \(t \in (-\infty,-2) \cup(-2,1)\cup(1,\infty)\text{.}\)

3. \(h(u) = \begin{cases} \dfrac{u^{2} - 1}{u-1} \amp u\neq 1\\ 2 \amp u = 1 \end{cases}\)

   Answer

   \(\left(-\infty,\infty\right)\)

   Solution

   \(\lim\limits_{u \to 1} h(u) = \lim\limits_{u \to 1} \dfrac{u^2 -1}{u-1} = \lim\limits_{u \to 1} \dfrac{(u-1)(u+1)}{(u-1)}= \lim\limits_{u \to 1} (u+1)=2\text{.}\) Since \(\lim\limits_{u \to 1} h(u) = 2 = h(1)\text{,}\) \(h\) is continuous at \(u=1\text{.}\) Additionally, we see that \(h\) is continuous everywhere else. Hence, \(h\) is continuous on \(\R\text{.}\)

Exercise 3.7.3.

Consider the function

\begin{equation*} h(x) = \left\{ \begin{array}{rl} 2x-3, \amp \mbox{if \(x\lt 1\),} \\ 0, \amp \mbox{if \(x\geq 1\).} \end{array} \right. \end{equation*}

Show that it is continuous at the point \(x=0\text{.}\) Is \(h\) a continuous function?

Answer Solution

Since \(\lim\limits_{x \to 0} h(x) = \lim\limits_{x \to 0} (2x-3)= -3=h(0)\text{,}\) \(h\) is continuous at \(x=0\text{.}\) However,

\begin{equation*} \lim\limits_{x \to 1^-} h(x) = \lim\limits_{x \to 1^-} (2x-3) =-1\text{,} \end{equation*}

and

\begin{equation*} \lim\limits_{x \to 1^+} h(x) = \lim\limits_{x \to 1^+} (0) = 0\text{,} \end{equation*}

and so \(\lim\limits_{x \to 1} h(x)\) DNE. Hence, \(h(x)\) is not continuous at \(x=1\) and is not a continuous function.

Exercise 3.7.4.

Find the values of \(a\) that make the function \(f(x)\) continuous for all real numbers.

\begin{equation*} f(x) = \left\{ \begin{array}{rl} 4x+5, \amp \mbox{if \(x\geq -2\),} \\ x^2+a, \amp \mbox{if \(x\lt -2\).} \end{array} \right. \end{equation*}

Answer Solution

First, we note that, for \(x > -2\text{,}\) \(f(x) = 4x+5\) is continuous. For \(x \lt -2\text{,}\) \(f(x) = x^{2} + a\) will be continuous for all choices of \(a\text{.}\) We now look at \(x = 2\text{.}\)

\begin{equation*} \begin{split} \lim_{x \to -2^{+}} f(x) \amp = \lim_{x \to -2^{+}} (4x + 5) = -3 \\ \lim_{x \to -2^{-}} f(x) \amp = \lim_{x \to -2^{-}} (x^{2} + a) = 4 + a \end{split} \end{equation*}

In order for the limit to exist, we therefore require \(-3 = 4 + a \implies a\) has to be \(-7\text{.}\) With this choice of \(a\text{,}\) we get that

\begin{equation*} \lim_{x \to -2} f(x) = -3 = f(-2)\text{,} \end{equation*}

and so \(f\) would be continuous. For all other choices of \(a\text{,}\) \(f\) would be discontinuous at \(x = -2\text{.}\)

Exercise 3.7.5.

Find the values of the constant \(c\) so that the function \(g(x)\) is continuous on \((-\infty,\infty)\text{,}\) where

\begin{equation*} g(x) = \left\{ \begin{array}{rl} 2-2c^2x, \amp \mbox{if \(x\lt -1\),} \\ 6-7cx^2, \amp \mbox{if \(x\geq-1\).} \end{array} \right. \end{equation*}

Answer Solution

First notice that \(g(x)\) is continuous for \(x \lt -1\) and \(x > -1\) since the branches of \(g\) are polynomial functions. Hence, we need only ensure continuity at \(x=-1\text{.}\) We compute the one-sided limits as \(x\) approaches \(-1\text{:}\)

\begin{equation*} \lim_{x\to -1^+} g(x) = \lim_{x\to -1^+} (6-7cx^2) = 6-7c, \text{ and } \end{equation*}

\begin{equation*} \lim_{x\to -1^-} g(x) = \lim_{x\to -1^-} (2-2c^2x) = 2+2c^2\text{.} \end{equation*}

For continuity, we require \(\lim\limits_{x\to -1^+} g(x) = \lim\limits_{x\to -1^-} g(x) = g(-1)\text{:}\)

\begin{equation*} 6-7c = 2+2c^2 \implies 2c^2+7c-4 = 0 \implies c=-4,\frac{1}{2}\text{.} \end{equation*}

Hence, \(g(x)\) is continuous for all \(x \in \mathbb{R}\) if \(c = -4\) or \(c= \frac{1}{2}\text{.}\)

Exercise 3.7.6.

A data plan at an internet café charges $1.00 for the first minute and $0.50 for each additional minute or part thereof, subject to a maximum of $3.00. Derive a function \(f\) relating the data charges to the length of time \(x\) spent at the café. Sketch the graph of \(f\) and determine the values of \(x\) for which the function is discontinuous.

Answer

At every minute (on the minute) after the first, the fee jumps $\(0.50\text{.}\) \(f\) is therefore discontinous at \(x = 1,2,..,5\text{.}\) After \(x =5\) minutes, the fee is constant and so \(f\) is continuous.

Exercise 3.7.7.

Approximate a root of \(\ds f=x^3-4x^2+2x+2\) to one decimal place.

Answer Solution

We first notice that \(f(0)=2\) and \(f(2)=-2\text{.}\) So \(f\) must have a root \(x \in (0,2)\) by the IVT. We construct the following table:

\begin{equation*} \begin{array}{c|c} x \amp f(x) \\ \hline 1 \amp 1 \\ 1.1 \amp 0.691 \\ 1.2 \amp 0.361 \\ 1.3 \amp 0.037 \\ 1.4 \amp -0.296 \end{array} \end{equation*}

Hence, \(f\) must have a root \(x \in (1.3, 1.4)\text{.}\) Since \(f(1.31) > 0\) and \(f(1.32)\lt 0\text{,}\) we have that \(x=1.3\) approximates a root of \(f\) to one decimal place.

Exercise 3.7.8.

Approximate a root of \(\ds f=x^4+x^3-5x+1\) to one decimal place.

Answer Solution

The given function \(f\) is continuous on \((-\infty,\infty)\) and so we can use the Intermediate Value Theorem. We start by constructing the following table:

\begin{equation*} \begin{array}{l|l} x \amp f(x) \\ \hline 0 \amp 1\\ 0.1 \amp 0.5011\\ 0.2 \amp 0.0096\\ 0.3 \amp -0.4649 \end{array} \end{equation*}

Therefore, there is a root in the interval \((0.2,0.3)\text{.}\) We finally compute that \(f(0.21) \lt 0\text{,}\) and so there is a root in the interval \((0.2,0.21)\text{.}\) To one decimal place, we can estimate a root of \(f\) to be \(0.2\text{.}\)

Exercise 3.7.9.

Show that the equation \(\sqrt[3]{x}+x=1\) has a solution in the interval \((0,8)\text{.}\)

Solution

Let \(f(x) = \sqrt[3]{x}+x-1\text{.}\) Since \(f(0)=-1\lt 0\) and \(f(8)=2+8-1>0\text{,}\) by the IVT, \(f\) must have a root in the interval \((0,8)\text{.}\) Since

\begin{equation*} f(x) = 0 \Leftrightarrow \sqrt[3]{x}+x=1\text{,} \end{equation*}

the equation must have a solution in \((0,8)\text{.}\)

martinezprefte.blogspot.com

Source: https://www.sfu.ca/math-coursenotes/Math%20157%20Course%20Notes/sec_ContinuityIVT.html

Share this post