Section 4.4 The Mean Value Theorem
The Mean Value Theorem is the intuitive fact that the slope between the endpoints of a curve is a mean (average) of the slopes at the various points along that curve, and so is between the extreme values of the tangent slopes, so that this secant slope equals the tangent slope at at least one point. A few details are needed to make this precise:
Theorem 4.4.1. The Mean Value Theorem, or MVT.
If a function \(f\) is continuous on a closed interval \([a,b]\) and is also differentiable there except possibly at the endpoints, then for at least one number \(c\) in \((a,b)\text{,}\) the average slope over the whole interval equals the tangent slope:
\begin{equation*}
f'(c) = \frac{f(b)-f(a)}{b-a}
\end{equation*}
That is, the change \(\Delta x=b-a\) over the interval produces a change
\begin{equation*}
f(b)-f(a) = f'(c)(b-a)
\mbox{, or }
\Delta y = f'(c)\Delta x.
\end{equation*}
We will verify the MVT for all cases soon, but for now:
Checkpoint 4.4.2.
Verify the Mean Value Theorem for \(f(x)=x^3-x\text{,}\) \(a=1\text{,}\) \(b=3\text{.}\)
The Case of Zero Mean Slope: Rolle's Theorem.
The MVT is even more intuitive in the special case when the value at each endpoint is the same, so the secant line is horizontal: the MVT then says that \(f'(c)=0\) somewhere in between.
Theorem 4.4.3. Rolle's Theorem.
If a function \(f\) is continuous on a closed interval \([a,b]\) and differentiable there except possibly at the endpoints, and if \(f(a)=f(b)\text{,}\) then \(f'(c)=0\) for at least one number \(c\) in the open interval \((a,b)\text{.}\)
This has to be true basically because
\(f\) must have a global maximum or minimum at some point
\(c\) between
\(a\) and
\(b\text{,}\) and then Fermat's Theorem in
Section 4.1 says that
\(f'(c)=0\text{.}\) The only way that Fermat's Theorem might fail to give
\(f'(c)=0\) is that the maximum and minimum both occur at the endpoints. But then the common endpoint value is both the global maximum and the minimum so
\(f\) is constant, making
\(f'(c)=0\) for any
\(c\text{!}\)
Example 4.4.4.
If an object's position \(s\) is a differentiable function of time \(t\text{,}\) \(s=f(t)\text{,}\) and the object is at the same position at two different times \(a\) and \(b\text{,}\) then Rolle's Theorem shows that the velocity \(v=s'\) is zero at some intermediate time: to return to its starting point, an object moving on a line must be stationary at some intermediate time.
Note that this confirms an intuition already used in
Section 3.7.
Checkpoint 4.4.5.
Prove that equation \(x^3+x-1=0\) has exactly one solution. Use the Intermediate Value Theorem to show that there is at least one solution, and then Rolle's Theorem to show that there is not more than one.
Average Velocity.
An intuitive application is that the average velocity over an interval of time must equal the instantaneous velocity at at least one moment: your speed cannot always be above average or always below average, and to swap between above and below, it must at some moment be exactly average. This comes from letting distance traveled by time \(t\) be \(s=f(t)\) so the average velocity between times \(a\) and \(b\) is \(\displaystyle v_{ave}=\frac{f(b)-f(a)}{b-a}\text{,}\) and at some intermediate time \(t=c\text{,}\) the instantaneous velocity is \(v_{inst}=f'(c)=v_{ave}\text{.}\)
Checkpoint 4.4.6.
Suppose that \(f(0)=-3\) and \(f'(x) \leq 5\) for all \(x\text{.}\) How large can \(f(2)\) possibly be?
Verifying the MVT in all cases.
It can now be seen that the MVT is true in all cases by simply “twisting” the graph to get a horizontal secant line. For \(f\) satisfying the conditions of the MVT and with \(m=\displaystyle \frac{f(b)-f(a)}{b-a}\) its mean slope, the new function
\begin{equation*}
g(x)=f(x)-m(x-a)
\end{equation*}
has the same value \(f(a)\) at each end point, and Rolle's Theorem gives \(g'(c)=0\) somewhere in between. Since \(f(x)=g(x)+m(x-a)\text{,}\) we get the claimed result that
\begin{equation*}
f'(c)=g'(c)+m=0+m=\frac{f(b)-f(a)}{b-a}\text{.}
\end{equation*}
Zero Derivative Means Constant on an Interval.
The MVT also helps us to find a function from information about its rate of change. For now, we deal with just the simplest case of zero rate of change:
Theorem 4.4.7.
If a function has derivative equal to zero at every point of an open interval \((a,b)\text{,}\) it is constant on that interval. This also applies to infinite intervals like \((-\infty,\infty)\text{.}\) That is, on intervals, “zero derivative everywhere” means “constant everywhere”.
Proof.
This is true because if the function were not constant, there would be a pair of numbers \(\alpha < \beta\) in the interval with \(f(\alpha) \neq f(\beta)\text{,}\) and then the MVT applied to this smaller closed interval \([\alpha,\beta]\) would give a number \(c\) with derivative \(f'(c) = (f(\beta)-f(\alpha)/(\beta-\alpha) \neq 0\text{,}\) which would contradict what we know about \(f'\text{.}\) Getting \(f\) from \(f'\text{,}\) part I.
The above says that the obvious functions with derivative \(f'=0\) are the only functions with this derivative, at least with domain being an interval. This has another important consequence:
Theorem 4.4.8.
If two functions have equal derivatives at each number in an interval, they differ by a constant on that interval.
That is, if \(f'(x)=g'(x)\) for all \(x\) in \((a,b)\text{,}\) then for some constant \(C\text{,}\) \(f(x)=g(x)+C\text{.}\)
Checkpoint 4.4.9.
(a) Find every function whose derivative is \(\cos x\text{.}\)
(b) Find every function whose derivative is \(1/x\text{.}\) Be careful!
Checkpoint 4.4.10.
Verify the identity \(\arcsin x + \arccos x = \pi/2\text{.}\) Do this two ways:
With a diagram and trigonometry.
Exercises Exercises
Study
Calculus Volume 1, Section 4.4 3 . Pay particular attention the
Corollaries of the Mean Value Theorem in the second half: Theorems 6, 7 and 8: these will be extremely useful for applications later in this chapter.
Study Examples 14 and 15, Checkpoint 14, and a selection from Exercises 148–150, 152–156, 161–166, 167–169, 182–184, and 190–193.
Here I group the exercises in ranges by "question type", so start by trying one or two from each of the six ranges. For example, some suggested selections are Exercises 149, 153, 161, 169, 182 and 192.
openstax.org/books/calculus-volume-1/pages/4-4-the-mean-value-theorem
openstax.org/books/calculus-volume-1/4-4-the-mean-value-theorem
openstax.org/books/calculus-volume-1/pages/4-4-the-mean-value-theorem
