7.1 Chains of Theorems

Appendix

Here we show that LUBA \(\pmb{\Longrightarrow}\) IVT, and IVT and MVT \(\pmb{\Longrightarrow}\) DT. The proofs can be omitted on a first reading, as they will not be used in subsequent sections.

LUBA \(\pmb{\Longrightarrow}\) IVT.

Proof Assume that \(f(a) < N < f(b)\), (the case \(f(b) < N < f(a)\) can be proven in a similar fashion, see ), and let

\[S= \{x\in [a,b]\,:\, f(x) \leq N\}.\]

Then, \(S\) has the following properties:

• – \(S\neq \emptyset\). This follows because \(f(a) < N\), thus, at least \(a\in S\).
• – \(S\) is bounded from above. This is clearly so as \(S\subset [a,b]\), thus \(x\leq b\) for all \(x \in S\).

Hence, by the LUBA, there exists a unique least upper bound \(c= \sup S\) of \(S\). We now show that \(f(c) = N\).

Recall that \(c= \sup S\) means that if \(x\in S\), then \(x \leq c\), and that if \(c_1\) is an upper bound of \(S\), then \(c \leq c_1\). Thus, in particular,

\(\hphantom{(*)}\) for any \(c_ 1< c\) there exists \(x\in S\) such that \(c_1 \leq x \leq c\).\((*)\)

There are three options for \(f(c)\): (i) \(f(c) < N\); (ii) \(f(c) > N\); (iii) \(f(c) = N\). We shall see that (i) and (ii) are impossible.

(i) If \(f(c) < N\), then, because \(f\) is continuous, it follows that for any \(\epsilon\) such that \( 0 < \epsilon < N-f(c)\) there exists a \(0 < \delta \), sufficiently small, such that

if c – δ < x < c + δ \( \Longrightarrow \) f(c) – ε < f(x) < f(c) + ε < f(c) + (N - f(c)) = N. \( \Longrightarrow \) [c,c + δ) \( \subset \) S.

See Figure (a). But this contradicts the fact that \(c\) is the least upper bound of \(S\). It follows that \(f(c) < N\) is impossible.

Figure 7.2: (a)

Figure 7.2: (b)

(ii) If \(f(c) > N\), then, by continuity, for any \( 0 < \epsilon < f(c) - N \) there exists a \(0 < \delta \), sufficiently small, such that

\[\scriptsize{ \text{if } \; c- \delta < x < c + \delta \;\; \Longrightarrow \;\; N < f(c)-\epsilon < f(x) < f(c) + \epsilon. \;\;\Longrightarrow \;\; x \notin S. }\]

See Figure (b). But this contradicts \((*)\) (all points \( c -\delta < x < c\) do not belong to \(S\)). Thus, \(f(c) > N\) is impossible.

As (i) and (ii) are ruled out, then (iii) is the only valid option, that is, \(f(c)=N\).

IVT and MVT \(\pmb{\Longrightarrow}\) DT.

Proof (The proof below appeared in Lars Olsen (2004) A New Proof of Darboux’s Theorem, The American Mathematical Monthly, 111:8, 713-715, DOI: 10.1080/00029890.2004.11920134.)

We can assume that \(N\) is between \(f'(a)\) and \(f'(b)\), and that \(N \neq f'(a)\), \(N \neq f'(b)\), and \(N \neq \frac{f(b)-f(a)}{b-a}\) (in the last case, the result will follow immediately from the MVT).

Define continuous functions \(f_a,\, f_b : I \to \mathbb{R}\) by

\[\scriptsize{f_a(x) =
\left \{\renewcommand{\arraystretch}{1.2}
\begin{array}{ll}
f'(a) & t = a\\
\displaystyle \frac{f(a)-f(t)}{a-t} & t \neq a\\
\end{array}\right .
\;\; \text{ and } \;\;
f_b(x) =
\left \{\renewcommand{\arraystretch}{1.2}
\begin{array}{ll}
f'(b) & t = b\\
\displaystyle \frac{f(t)-f(b)}{t-b} & t \neq b\\
\end{array}\right .}\]

Note that

\[f_a(a) = f'(a), \;\; f_a(b) = f_b(a), \;\; f_b(b) = f'(b).\]

Thus, (i) \(N\) is between \(f_a(a)\) and \(f_a(b)\), or (ii) \(N\) is between \(f_b(a)\) and \(f_b(b)\).

In the first case, since \(f_a\) is continuous, there exists \(d\in (a,b)\) such that

\[f_a(d) = N \Longleftrightarrow N =\frac{f(d)-f(b)}{d-b}.\]

The MVT then implies that there exists \(c\in (d,b)\) such that

\[N = \frac{f(d)-f(b)}{d-b} = f'(c).\]

7.1.4 Exercises

1. Prove that if, in the notation of FT, \(f(c)\) is an absolute minimum of \(f\), then \(f'(c)=0\).
2. Complete the details of DT, for the case (ii) (see the proof above).
3. Write detailed proofs of the following statements:

   1. EVT & FT \(\Longrightarrow\) RT;
   2. RT \(\Longrightarrow\) MVT.
   3. RT \(\Longrightarrow\) CMVT.
4. Complete the proof of the IVT by providing all details to prove the case \(f(b) < N < f(a)\).
5. For each set \(S\), (i) show that the set is bounded from above; (ii) propose a least upper bound \(\sup S\); (iii) verify that \(\sup S\) is an upper bound for \(S\) and that any other upper bound \(M\) for \(S\) satisfies \(\sup S \leq M\).
1. \(\displaystyle S = \left \{1- \frac{1}{n}\, : \, n \in \mathbb{N} \right \}\);
2. \(\displaystyle S = \left \{\arctan n \, : \, n \in \mathbb{N} \right \}\);
3. \(\displaystyle S = \left \{ x\in \mathbb{Q} \, : \, x^2 < 2 \right \}\);
4. \(\displaystyle S = \left \{ 1 – e^{-n} \, : \, n \in \mathbb{N} \right \}\).

7.1.5 Existence Theorems

The Intermediate Value Theorem (IVT), the Extreme Value Theorem (EVT), Rolle’s Theorem (RT), and the Mean Value Theorem (MVT) are existence theorems. That is, the conclusion of each one of these theorems guarantees the existence of a certain real number \(c\) satisfying some properties, but, none of these theorems provides an explicit method to find such a number \(c\), nor determines if such a number is unique of if there are more such numbers than just one. However, in many instances, such an existence is all that is needed to obtain a desired result. As examples of this, we present (i) existence of roots of equations; (ii) existence of extreme values of a function; (iii) existence of horizontal tangent lines; (iv) existence of tangent lines parallel to secant lines to the graph of a function; (v) estimates of the distance between the values of a function; (vi) an integral formula to compute the length of the graph of a function. (vii) We conclude the section by showing how, in some cases, it is possible to combine the IVT and RT to show the existence and uniqueness of roots of equations.

Existence of Roots of Equations}

Example 1 Let \(f(x) =x^3 – 2 x^2 – 3 x + 3\), \(x\in [-2,3]\). Show that the equation \(f(x)= -1\) has at least one root on the interval \([-2,2]\).

Solution. If, in the notation of the IVT (see Theorem in ), we let \(N=-1\), then, it is immediate that \(f\) satisfies the hypothesis of the theorem. Since \(f\) is a polynomial function, it is continuous every where, and since \(f(-2) = -7\) and \(f(3) = 3\), then \(f(-2) < N < f(3)\). Thus, by the IVT, there exists a number \(c\in (-2,3)\) such that \(f(c)= -1\). That is, there exists at least one root of the equation \(x^3 - 2 x^2 - 3 x + 3 =-1\) on the interval \((-2,3)\).

The Fundamental Theorem of Algebra (also an existence theorem) guarantees that there exist three solutions (counting multiplicities) of the equation \(f(x)= -1\) . What about the other two? As it turns out, the graph of \(f\) shows that the three roots of the equation can be found on the interval \((-2,3)\) (see Figure (a)).

Extreme Values and Horizontal Tangent Lines

Example 2 Let \(f(x) = 4 (x^4 – x^2)\sin \pi x\), \(x\in [-1,1]\). Then, \(f\) clearly satisfies the hypothesis of the EVT. It follows that there exists \(c_1, \, c_2 \in [-1,1]\) such that \(f(c_1)\) is an absolute maximum and \(f(c_2)\) is an absolute minimum of \(f\) on \([-1,1]\). Figure (b) shows these highest and lowest points on the graph of \(f\). However, it is not immediate how to find the exact value of \(c_1\) and \( c_2\).

Figure 7.3: (a) IVT

Figure 7.3: (b) EVT & RT

Figure 7.3: (c) MVT

Example 3 (Continuation of Example .) Let \(f(x) = 4 (x^4 – x^2)\sin \pi x\), \(x\in [-1,1]\). It is immediate that \(f\) satisfies the assumptions of RT. Thus, there is a \(c\in (-1,1)\) such that \(f'(c)=0\). The proof of RT (see ) shows that we can take \(c=c_1\) or \(c=c_2\) (as in Example , see Figure (b)). However, once again, it is not possible, from this information, to determine the exact value of \(c\).

Tangent Lines Parallel to Secant Lines

Example 4 Consider the function

\[\scriptsize{f(x)=-\frac{7 x^6}{120}+\frac{47 x^5}{120}-\frac{7 x^4}{24}-\frac{17 x^3}{8}+\frac{47 x^2}{20}+\frac{41 x}{15}-1,\;\;x\in [-2,3].}\]

It is clear that \(f\) satisfies the hypothesis of the MVT. Thus, there exists a number \(c\in (-2,3)\) such that

\[ \hphantom{(*)} f'(c) = \frac{f(3)-f(-2)}{3-(-2)}. \hspace{1.5in} (*)\]

Figure (c) shows a tangent line parallel to the secant line passing through the points \(A(-2,f(-2))\) and \(B(3,f(3))\). However, the theorem does not provide any additional information about the precise value of \(c\) or if \(c\) is unique.

In fact, the graph intuitively shows that there exist three more possible values of \(c\) for which \((*)\) is satisfied (explain).

Comparing the Values of a Function and the MVT

Example 5 Show that \(|\sin x – \sin y| \leq |x-y|\) for all \(x,\, y \in \mathbb{R}\).

Solution. Let \(x,\, y \in \mathbb{R}\) with \(x < y\) (the case \(y < x\) is proven similarly). Then, \(f(x) = \sin x\) satisfies the assumptions of the MVT on \([x,y]\). It follows that there exists \(c\in (x,y)\) such that

\[\scriptsize{\sin x – \sin y = f'(c)(x-y) \;\; \Longrightarrow \;\; |\sin x – \sin y | =| f'(c)(x-y) |\;\; \Longrightarrow }\]

\[\scriptsize{\Longrightarrow \;\; |\sin x – \sin y | =| \cos(c)||x-y |\;\;\Longrightarrow \;\; |\sin x – \sin y | \leq |x-y|.}\]

Where we have used, in the last inequality, the fact that \(|cos c| \leq 1\) for all \(c\in \mathbb{R}\).

Example 6 Show that \(e^x > x+1\) for all \(x > 0\).

Solution. If we apply the MVT to \(f(x)=e^x\) over the interval \([0,x]\), for \(x > 0\), we have that there exists a \(c\in (0,x)\) such that

\[\small{\frac{f(x) – f(0)}{x-0} = f'(c) \Longrightarrow e^x-1 = e^c(x-0) \Longleftrightarrow e^x = e^c x + 1.}\]

Since \(e^c > 1\) for all \(c > 0\), we get \(e^x > x + 1\) for all \(x > 0\).

Computing the Arc Length of the Graph of a Function and the MVT

Integral Formula to Compute Arc Length

Let \(f\) be a function on the closed interval \(I = [a,b]\) with continuous first derivative on \(I\). Then, the arc length of the graph of \(f\) on \([a,b]\) is given by

\begin{equation}
L = \int_a^b\sqrt{1+f'(x)}\, dx.
\end{equation}

Proof. Let \(a=x_0, x_1, \dots , x_n=b\) be the endpoints of a partition of \(I\) into \(n\) subintervals. For each subinterval \([x_{i-1},x_i]\), \(i=1, \dots , n\), we compute the length \(\ell_i\) of the \(i\)-th line segment connecting the points \(P_{i-1}(x_{i-1}, f(x_{i-1}))\) and \(P_{i}(x_i, f(x_i))\) on the graph of \(f\):

\[\ell_i = \sqrt{\left (x_i – x_{i-1}\right )^2 +\left (f(x_i) – f(x_{i-1})\right )^2 }, \;\; i = 1, 2, \dots , n\]

(see Figure ).

Figure 7.4: Polygonal approximation to the graph of f

Then, we have,

\[L \approx \sum_{i=1}^n \ell_i = \sum_{i=1}^n \sqrt{\left (x_i – x_{i-1}\right )^2 +\left (f(x_i) – f(x_{i-1})\right )^2 }.\]

We are going to show that the sum on the right-hand-side is a Riemann sum, and that its limit is precisely the integral formula in (). To accomplish this, we use the MVT.

Note that \(f\) satisfies the assumptions of the MVT on each subinterval \([x_{i-1},x_i]\), \(i=1, \dots , n\), thus, for each such interval, there exists a point \(c_i\in (x_{i-1},x_i)\) such that

\[\small{f'(c_i)=\frac{f(x_i) – f'(x_{i-1})}{x_i – x_{i-1}}\;\;\text{ or } \;\; f(x_i) – f(x_{i-1}) = f'(c_i)(x_i – x_{i-1}).}\]

Substituting these expressions for \(f(x_i) – f(x_{i-1}) \) into the approximation to \(L\) above, we get

\[ \small{\begin{array}{cl}
L \approx \sum_{i=1}^n \ell_i &= \sum_{i=1}^n \sqrt{\left (x_i – x_{i-1}\right )^2 +\left ( f'(c_i)(x_i – x_{i-1})\right )^2 }
\\&=\sum_{i=1}^n \sqrt{\left (x_i – x_{i-1}\right )^2\left (1 +\left ( f'(c_i)\right )^2\right ) } \\
&= \sum_{i=1}^n \sqrt{1 +\left ( f'(c_i)\right )^2 }\,\left (x_i – x_{i-1}\right )
\\&= \sum_{i=1}^n \sqrt{1 +\left ( f'(c_i)\right )^2 }\,\Delta x_i.
\end{array}}\]

Here, \(\Delta x_i\) denotes the length of the \(i\)-th subinterval. This sum
is a Riemann sum. Recall that (see ) the limit with \(n\) increasing indefinitely and \(\|\Delta x\| \to 0\) at the same time,

\[ \lim_{n\to \infty, \; ||\Delta x|| \to 0}
\sum_{i=1}^{n} \sqrt{1 +\left ( f'(c_i)\right )^2 }\,\Delta x_i,\]

if it exists, is the definite integral as in ().
Furthermore, the limit exists if \(\sqrt{1 +\left ( f'(x)\right )^2 }\) is continuous on \(I\) (hence our initial assumption that \(f’\) be continuous on \(I\)).

Example 7 Compute the length of the graph of \(f(x) = \ln \sin x \), \(\displaystyle x \in \left [\frac{\pi}{6}, \frac{\pi}{3}\right ]\).

Solution. Since \(\displaystyle f'(x) = \frac{1}{\sin x}\cos x = \cot x\), \(f’\) is continuous on \(\displaystyle \left [\frac{\pi}{6}, \frac{\pi}{3}\right ]\). A direct substitution into () yields

\[L = \int_{\pi/6}^{\pi/3}\sqrt{1+ \cot^2x}\,dx = \int_{\pi/6}^{\pi/3}\csc x\,dx .\]

To find an antiderivative of \(g(x) = \csc x\), proceed as follows:

\[\int \csc x \,dx
= \int \csc x\cdot \frac{\csc x + \cot x}{\csc x + \cot x} \,dx\]

\[= \int \frac{\csc^2 x + \csc x \cot x}{\csc x + \cot x} \,dx .\]

Note now that if \(h(x) = \csc x + \cot x \), then \(h'(x) =-(\csc x \cot x + \csc^2 x)\), thus

\[\int \csc x \,dx =-\ln |\csc x + \cot x |+C.\]

Hence

\[ \begin{array}{cl}L &= -\ln |\csc x + \cot x |\Big|_{\pi/6}^{\pi/3} \\
&= -\ln\left |\csc \frac{\pi}{3} + \cot \frac{\pi}{3} \right | + \ln \left |\csc \frac{\pi}{6} + \cot \frac{\pi}{6} \right | \\
&= \ln \frac{\sqrt{3}+2 }{\sqrt{3}} .\end{array} \]

Existence and Uniqueness: IVT and RT

In some cases, it is possible to combine the Intermediate Value Theorem and Rolle’s Theorem to give more information about the existence and the number of solutions of an equation.

Example 8 Show that the equation \(3x – 2\sin x =1\) has a unique root.

Solution. First, we prove that the equation has at least one solution. For this, we use the IVT with (in the notation of Theorem in \(f(x) = 3x – 2\sin x \), and \(N=1\). We need to find an interval \([a,b]\) where \(f\) is continuous and \(N\) lies between \(f(a)\) and \(f(b)\). We do this by trial and error, choosing values for which it is easy to evaluate \(f\). For Example, \(f(0) = 0\) and \(f(\pi) = 3\pi\). Since, \(f(0) < 1 < f(\pi)\), we can choose as a closed interval \(\displaystyle \left [0, \pi\right ]\). Since \(f\) is continuous everywhere, in particular, it is continuous on \(\displaystyle \left [0, \pi\right ]\). Thus, \(f\) satisfies the assumptions of the IVT. Hence, there exists \(c\in (0,\pi)\) such that

\[f(c)= 1\;\; \Longrightarrow \;\; 3c – 2\sin c =1.\]

We now need to prove that the solution is unique. We will do this by contradiction. Assume that there exists a second root of the equation \(c_0\). That is, assume that we have two solutions

\[f(c)=1\;\; \text{and}\;\; f(c_0) = 1.\]

Assuming that \(c < c_0\) (the other case can be treated similarly), we now note that \(f\) satisfies the hypothesis of RT on the interval \([c, c_0]\). (i) Clearly, \(f\) is continuous on \([c, c_0]\); (ii) \(f\) is differentiable on \((c, c_0)\). In fact, \(f'(x) = 3-2\cos x\) for all \(x\in \mathbb{R}\); (iii) \(f(c) = f(c_0)\) (both are equal to \(1\)). It follows that there exists at least one \(s\in (c, c_0)\) such that \(f'(s) = 0\). However, this is impossible, since

\[f'(s) = 3-2\cos s > 0\;\; \forall \; s \in \mathbb{R}.\]

Thus, our assumption that the equation \(3x – 2\sin x =1\) has two roots is wrong. And the equation has a unique solution.

7.1.6 Exercises

1. Apply the IVT to show that the equation has at least one root. State the interval.

   (a) \( x^5 – 3 x^2 – x + 4 = 0\);
   (b) \(x \sin x = 2x-1\);
   (c) \(x^2\arcsin x = 2x-1\).

2. (i) Show that \(f\) satisfies the hypothesis of the
   MVT on the given interval \(I\); (ii) find all numbers \(c\) that satisfy the conclusion of the MVT.

   (a) \(f(x) = x^3 + 3 x\), \(I=[2,4]\);
   (b) \(\displaystyle f(x) = \frac{x+1}{2x+1}\), \(I=[1,2]\);
   (c) \(\displaystyle f(x) = \frac{x^2-3}{x-2}\), \(I=[-3,1]\).

3. Explain why the MVT cannot be applied for each function on the given interval.

   (a) \(f(x) = \tan \pi x\), \(\displaystyle I=\left [\frac{4}{3},2\right ]\);
   (b) \(\displaystyle f(x) = |2x-1|\), \(I=[0,1]\).

4. (i) Show that \(f\) satisfies the hypothesis of RT on the given interval \(I\); (ii) find all numbers \(c\) that satisfy the conclusion of RT.
   1. \(f(x) = \csc^2 2\pi x\), \(\displaystyle I=\left [-\frac{1}{3}, -\frac{1}{6}\right ]\);
   2. \(f(x) = \cos^2 2\pi x\), \(\displaystyle I=\left [-\frac{1}{6}, \frac{1}{3}\right ]\);
   3. \(f(x) = \tan^{-1}(x^3-x+1)\), \(I=[-1,1]\);
   4. \(\displaystyle f(x) = 3 x^5-5 x^3\), \(\displaystyle I=\left [-\sqrt{\frac{5}{3}},\sqrt{\frac{5}{3}}\right ]\);
   5. \(f(x) = \ln(x^2 e^{-x^2}+1)\), \(I=[-2,2]\);
   6. \(f(x) = (x – 2)^4 (x + 1)^2\), \(I=[0,3]\).

In problems – , prove the validity of each inequality.

5. \(|\tan x – \tan y| \geq |x -y| \) for all \(\displaystyle x, \, y \in \left (-\frac{\pi}{2}, \frac{\pi}{2}\right )\).
6. \(|\arctan x – \arctan y| \leq |x -y| \) for all \(\displaystyle x, \, y \in \mathbb{R}\).
7. \(|\cos x – \cos y| \leq |x -y| \) for all \(\displaystyle x, \, y \in \mathbb{R}\).

8. \(\displaystyle \ln \left (1 + \frac{1}{x}\right ) \leq \frac{1}{x} \) for all \(\displaystyle x > 0\).
9. 1. \(\displaystyle \arcsin x \geq x\) for all \(\displaystyle 0 < x < 1\);
   2. \(\displaystyle \arccos x \geq x\) for all \(\displaystyle 0 < x < 1\).

10. \(\displaystyle e^x > 1+ x + \frac{x^2}{2}\) for all \(\displaystyle x > 0\).
    Hint: see Example .

11. \(\displaystyle \left (1 +x\right )^n \leq 1 + n x\) for all \(\displaystyle x > 0\).

In problems – , compute the arc length of the graph of the function.

12. \(f(x) = \ln \cos x\), \(\displaystyle x\in \left [-\frac{\pi}{4}, \frac{\pi}{3}\right ]\).
13. \(f(x) = \sqrt{x^3}\), \(\displaystyle x\in \left [0,4\right ]\).
14. \(\displaystyle f(x) = \frac{x^3}{3}+\frac{1}{4x}\), \(\displaystyle x\in \left [1, 2\right ]\).
15. \(f(x) = \sqrt[3]{x^2}\), \(\displaystyle x\in \left [0,8\right ]\).
16. \(f(x) = \cosh x\), \(\displaystyle x\in \left [-\ln2, \ln 2\right ]\).
17. \(\displaystyle f(x) = \frac{x^4}{4}+\frac{1}{8x^2}\), \(\displaystyle x\in \left [1, 2\right ]\).

18. Let \(\displaystyle F(x)= \frac{1}{2}\left (x\sqrt{1+x^2} + \ln \left (\sqrt{1+x^2} + x \right )\right )\),
    1. show that \(F\) is an antiderivative of \(f(x) = \sqrt{1+x^2}\);
    2. compute the arc length of the graph of \(\displaystyle g(x)= \frac{1}{2}x^2\), \(x\in [0,1]\).
19. Show that the equation has exactly one real root on the given interval.
    1. \(2\sin x + 3 = 4 x \), \(\displaystyle x\in \left [ -\pi , \frac{3\pi}{2} \right ]\);
    2. \(\displaystyle xe^x = 1-2x \), \(\displaystyle x\in \left [ -1 ,1 \right ]\);
    3. \(\displaystyle 2^{2x+1} =4-4x \), \(\displaystyle x\in \left [0 ,\frac{1}{2} \right ]\);
    4. \(\displaystyle \ln x^3 = -2x+4 \), \(\displaystyle x\in \left [1 , 2\right ]\);
20. Show that the graphs of the functions \(y = x^x\) and \(y = -2x+6\) intersect at only one point.
21. Show that the equation \(x^5 + 20x^2+b=0\) has at most three real roots for any \(b\in \mathbb{R}\).
22. Prove that the equation has exactly two real roots.
    1. \( x^2=\cos x\);
    2. \( x^2 = x\sin x +\cos x\).