Objectives
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Understand basic theorems of functions, like the Intermediate Value Theorem, Rolle’s Theorem and the Mean Value Theorem.
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Use these theorems to demonstrate properties of functions.
Section 1.2 Important Theorems from Calculus
We now review a number of theorems that are usually presented in Calculus. Here we both state a number of important theorems, but show some examples and then prove most of these. For additional information on these, check any standard Calculus book. A nice open-source one can be found at APEX Calculus.
Theorem 1.2.1. Intermediate Value Theorem.
Let \(f\) be a continuous function on \([a,b]\text{.}\) Let \(M=\max(f(a),f(b))\) and \(m=\min(f(a),f(b))\text{.}\) The function \(f\) takes on all values between \(m\) and \(M\text{.}\) That is, for any value \(h\text{,}\) such that \(m\leq h\leq M\text{,}\) then there exists at least one value \(c\) such that \(f(c)=h\text{.}\)
This is a important theorem, however, there is much background needed to prove this. A good open-source book with a proof is in Introduction to Real Analysis by William F. Trench.
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The background and proof of this is in section 2.2
Theorem 1.2.2. Rolle’s Theorem.
Let \(f\) be a function that satisfies:
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\(f(a)=f(b)\text{.}\)
Proof.
From the Intermediate Value Theorem, \(f\) takes on a minimum value, \(m\) and maximum value, \(M\) on the interval \([a,b]\text{.}\)
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If \(m=M\text{,}\) then the function is constant and all values of \(c\) in the interval \((a,b)\) satisfy \(f'(c)=0\text{.}\)
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Otherwise, either the minimum or maximum or both of \(f\) occurs in \((a,b)\text{.}\) Let \(c\) be this value. Since the function is differentiable at \(c\) and is a minimum or a maximum, then \(f'(c)=0\text{.}\)
Example 1.2.3.
Show that \(f(x)=x^{2} +6x\) on the interval \([-6,0]\) satisfies all the conditions of Rolle’s Theorem. Find all of the values of \(c\) such that \(f'(c)=0\text{.}\)
Solution.
Since \(f\) is a quadratic, it is continuous and differentiable everywhere, so the first two conditions are satisfied. Also, \(f(-6)=(-6)^{2}+6(-6)=0\) and \(f(0)=0^{2}+6(0)=0\text{,}\) so the function is equal on the endpoints of the interval and the 3rd condition is satisfied.
To find the values of \(c\) such that \(f'(c)=0\text{,}\) we solve \(f'(c)=2c+6=0\) or \(c=-3\text{.}\)
A graphical view of this is as follows:
and one can see that the function is 0 at both \(x=-6\) and \(x=0\) and that there is a horizontal tangent line at \(c=-3\text{.}\)
We can extend Rolle’s theorem to functions that have more than 2 equal points. We will show an example of the theorem, but in general, this is very important for proving other theorems.
Theorem 1.2.5. Generalized Rolle’s Theorem.
If \(f\) is continuous on \([a,b]\text{,}\) has \(n\) continuous derivatives on \((a,b)\) and is equal to zero at \(n+1\) distinct points in \([a,b]\) then there exists a number \(c\in(a,b)\) with \(f^{(n)}(c) = 0\text{.}\)
Proof.
First, for some notation, assume that \(a_{k}\) is the \(k\)th zero of \(f\) for \(k=1,2,\ldots,n+1\) and the \(a_{k}\) are ordered increasingly. Therefore there are \(n\) subintervals of \([a,b]\text{,}\) denoted \([a_{k},a_{k+1}]\) such that \(f(a_{k})=f(a_{k+1})=0\text{.}\) Using Rolle’s theorem, for each subinterval, there exists a number \(c_{k}\) in \((a_{k},a_{k+1})\) such that \(f'(c_{k})=0\text{.}\)
Therefore there are \(n\) numbers, \(c_{k}\) such that the derivative is 0 or subintervals \([c_{k},c_{k+1}]\) such that \(f'(c_{k})=f'(c_{k+1})=0\text{.}\) Using Rolle’s theorem again, for each \(k\text{,}\) there exists a number \(d_{k} \in (c_{k},c_{k+1})\) such that \(f''(d_{k})=0\text{.}\)
Repeat this process \(n-2\) times and the theorem holds.
The following is a simple example of Generalized Rolle’s theorem. We will see this theorem used in a proof later in the text.
Example 1.2.6.
Let \(p(x)=(x+2)(x+1)x(x-1)(x-2)\text{.}\) Show that this satisfies Generalized Rolle’s theorem on \([-2,2]\) for \(n=4\) and find the value of \(c\) such that \(f^{(4)}(c)=0\text{.}\)
First, there are zeros of \(p\) at \(x=-2,-1,0,1,2\) and since \(p\) is a polynomial, it is infinitely differentiable, so it satisfies the conditions of the theorem.
Also, \(p\) can be written:
\begin{equation*}
p(x) = x(x^{2}-1)(x^{2}-4) = x^{5}-5x^{3}+4x
\end{equation*}
Also, consider a plot of the above situation.
and as can be seen that the function is 0 at the 5 points \(x=-2,-1,0,1,2\) and any function with 5 zeros like this either wiggle a lot like the above function, or is constant.
Theorem 1.2.8. Mean Value Theorem.
Let \(f\) be a function that is continuous on a closed interval \([a,b]\) and differentiable on the interval \((a,b)\text{.}\) Then there exists a number \(\xi \in (a,b)\) such that
\begin{equation*}
f'(\xi)= \frac{f(b)-f(a)}{b-a}
\end{equation*}
Proof.
Then
\begin{equation*}
g'(x) = f'(x) -\frac{f(b)-f(a)}{b-a}
\end{equation*}
and evaluated \(g\) on the endpoints of the interval shows that
\begin{equation*}
\begin{aligned} g(a) \amp = f(a) + (a-a) \frac{f(b)-f(a)}{b-a}- f(a) = 0, \\ g(b)\amp= f(b) + (b-a) \frac{f(b)-f(a)}{b-a}- f(a) = 0.
\end{aligned}
\end{equation*}
Since \(f\) is continuous on \([a,b]\) and differentiable of \((a,b)\text{,}\) so is \(g\) and therefore \(g\) satisfies Rolle’s Theorem. Then there exists a number \(c\) in the interval \((a,b)\) such that \(g'(c)=0\) or
\begin{align*}
g'(c) \amp = f'(c) - \frac{f(b)-f(a)}{b-a}= 0\\
\text{or} \qquad \amp \\
f'(c) \amp = \frac{f(b)-f(a)}{b-a}
\end{align*}
so the theorem holds.
The following example shows a nice consequence of the mean value theorem.
Example 1.2.9.
Solution.
Let \(f(x) = e^{x}\text{,}\) \(b=x\text{,}\) \(a=0\) then the MVT applied to this situation is
\begin{equation*}
e^{\xi} = \frac{e^{x} - 1}{x-0}
\end{equation*}
since \(e^{\xi}\geq 1\text{,}\) for \(\xi\geq 0\text{,}\) then
\begin{equation*}
\begin{aligned} \frac{e^{x}-1}{x}\amp \geq 1 \\ e^{x} \amp \geq x+1 \end{aligned}
\end{equation*}
