Section 3.7 Continuity and IVT
ΒΆSubsection 3.7.1 Continuity
ΒΆThe 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.Definition 3.54. Continuous at a Point.
A function f is continuous at a point a if
Guideline for Checking Continuity at a Point.
A function f(x) is continuous at x=a if the following three conditions hold:
f(a) is defined (that is, a belongs to the domain of f),
limxβaf(x) exists (that is, left-hand limit = right-hand limit),
limxβaf(x)=f(a) (that is, the numbers from 1 and 3 are equal).
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 (ββ,β)..
Summary of Discontinuities.
Example 3.56. Continuous at a Point.
What value of c will make the following function f(x) continuous at 2?
In order to be continuous at \(2\) we require
to hold. We use the three part definition listed previously to check this.
First, \(f(2)=c\text{,}\) and \(c\) is some real number. Thus, \(f(2)\) is defined.
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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 \(3\text{.}\)
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 \(2\text{.}\)
Definition 3.57. Continuous from the Right and from the Left.
A function f is left continuous at a point a if
and right continuous at a point a if
Definition 3.58. Continuity on a Closed Interval.
A function f is continuous on the closed interval [a,b] if:
it is continuous on the open interval (a,b);
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it is left continuous at point a:
limxβaβf(x)=f(a);and
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it is right continuous at point b:
limxβb+f(x)=f(b).
Example 3.59. Continuity on Other Intervals.
The function f(x)=βx is continuous on the (closed) interval [0,β).
The function f(x)=β4βx is continuous on the (closed) interval (ββ,4].
Theorem 3.60. Operations of Continuous Functions.
If f and g are continous at a, and c is a constant, then the following functions are also continuous at a:
fΒ±g;
cf;
fg;
f/g (provided g(a)β 0).
Example 3.61. Common Types of Continuous Functions.
Polynomials (for all x), e.g., y=mx+b, y=ax2+bx+c.
Rational functions (except at points x which gives division by zero).
Root functions nβx (for all x if n is odd, and for xβ₯0 if n is even).
Trigonometric functions
Inverse trigonometric functions
Exponential funtions
Logarithmic functions
Example 3.62. Evaluate a Limit.
Evaluate the following limit: limxβΟβx+sinx1+x+cosx.
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,
is a continuous function at \(\pi\) since \(1+\pi+\cos\pi\neq 0\text{.}\) Hence, we can directly substitute to get the limit:
Theorem 3.63. Continuity of Function Composition.
If g is continuous at a and f is continuous at g(a), then the composition function fβg is continuous at a.
Example 3.64. Continuity with Composition of Functions.
Determine where the following functions is continuous:
h(x)=cos(x2)
H(x)=ln(1+sin(x))
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.
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
ΒΆWhether 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.
Does ex+x2=0 have a solution?
Does ex+x=0 have a solution?
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.
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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 \(x\approx -0.56714\text{.}\)
Theorem 3.66. Intermediate Value Theorem.
If f is continuous on the interval [a,b] and N is between f(a) and f(b), where f(a)β f(b), then there is a number c in (a,b) such that f(c)=N.
Example 3.67. Intermediate Value Theorem.
Show that there is a solution of 3βx+x=1 in the interval (0,8).
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 f=x3+3x2+xβ2 has a root between 0 and 1.
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 \(f(c)=0\text{.}\)
Example 3.69. Approximating Roots.
Approximate the root of the previous example to one decimal place.
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.
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Answer
\(f(x) = \begin{cases} x - 2 \amp x \leq 0 \\ 2 \amp x > 0 \end{cases}\)
Solution\(x = 0\text{.}\) \(\lim\limits_{x \to 0}\) does not exist and is not equal to \(f(0)\text{.}\)
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 \(x = 0\text{.}\)
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Answer
\(f(x) = \begin{cases} x + 2 \amp x > 0 \\ -\frac{1}{2}x^{2}+2 \amp x \leq 0 \end{cases}\)
SolutionNo points of discontinuity. \(f\) is continuous everywhere.
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.
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Answer
\(f(x) = \begin{cases} x + 2 \amp x > 0 \\ -5 \amp x = 0 \\ -\frac{1}{2}x^{2}+2 \amp x > 0 \end{cases}\)
Solution\(x = 0\text{.}\) \(\lim\limits_{x \to 0} = 2\text{,}\) which is not equal to \(f(0) = 1\text{.}\)
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 \(x = 0\text{.}\)
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Answer
\(f(x) = \begin{cases} \ln(x+2) \amp x \leq 1, x \neq -1 \\ -1 \amp x = -1 \\ -1 \amp x > 1 \end{cases}\)
Solution\(x=\pm 1\text{.}\) \(\lim\limits_{x\to 1}\) DNE, \(\lim\limits_{x\to -1} \neq f(-1)\text{.}\)
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.
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\(f(s) = \dfrac{2}{s^{2} + 1}\)
AnswerSolution\(\left(-\infty,\infty\right)\)Since \(s^2+1=0\) has no real solutions, we see that \(f\) is continuous for all \(x \in (-\infty,\infty)\text{.}\)
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\(g(t) = \dfrac{2t+1}{t^{2}+t-2}\)
AnswerSolution\(\left(-\infty, -2 \right) \cup \left(-2,1\right) \cup \left(1,\infty\right)\)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 \(t \in (-\infty,-2) \cup(-2,1)\cup(1,\infty)\text{.}\)
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\(h(u) = \begin{cases} \dfrac{u^{2} - 1}{u-1} \amp u\neq 1\\ 2 \amp u = 1 \end{cases}\)
AnswerSolution\(\left(-\infty,\infty\right)\)\(\lim\limits_{u \to 1} h(u) = \lim\limits_{u \to 1} \dfrac{u^2 -1}{u-1} = \lim\limits_{u \to 1} \dfrac{(u-1)(u+1)}{(u-1)}= \lim\limits_{u \to 1} (u+1)=2\text{.}\) 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
Show that it is continuous at the point \(x=0\text{.}\) Is \(h\) a continuous function?
AnswerSince \(\lim\limits_{x \to 0} h(x) = \lim\limits_{x \to 0} (2x-3)= -3=h(0)\text{,}\) \(h\) is continuous at \(x=0\text{.}\) However,
and
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.
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 \(x = 2\text{.}\)
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
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
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 \(-1\text{:}\)
For continuity, we require \(\lim\limits_{x\to -1^+} g(x) = \lim\limits_{x\to -1^-} g(x) = g(-1)\text{:}\)
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 cafΓ© 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 cafΓ©. Sketch the graph of \(f\) and determine the values of \(x\) for which the function is discontinuous.
AnswerAt 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.
AnswerWe 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:
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.
AnswerThe given function \(f\) is continuous on \((-\infty,\infty)\) and so we can use the Intermediate Value Theorem. We start by constructing the following table:
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 \((0,8)\text{.}\)
SolutionLet \(f(x) = \sqrt[3]{x}+x-1\text{.}\) 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
the equation must have a solution in \((0,8)\text{.}\)