Section 1.4 The Mean Value Theorem
¶Motivating Questions
How can we use the size of the derivative to quantify how fast a function is increasing or decreasing?
In Active Calculus 3.1 we learned about critical points and how to use them to find relative maximums and relative minimums. Without giving it a name, we used the following theorem.
Theorem 1.4.1 Fermat's Theorem
If \(f\) has a relative maximimum or relative minimum at \(c\text{,}\)
then either \(f'(c)=0\) or \(f'(c)\) does not exist.
In Active Calculus 3.3 we learned the following theorem, which we state now in slightly different form.
Theorem 1.4.2 The Extreme Value Theorem
If \(f\) is continuous on a closed interval \([a,b]\text{,}\)
then \(f\) attains both an absolute maximum and an absolute minimum on \([a,b]\text{.}\)
Preview Activity 1.4.1
For each of the following cases, either draw a function that satisfies the conditions or explain why it is impossible.
- \(f'(c)=0\) and \(f\) has a relative minimum at \(c\text{.}\)
- \(f'(c)=1\) and \(f\) has a relative minimum at \(c\text{.}\)
- \(f'(c)\) does not exist and \(f\) has a relative minimum at \(c\text{.}\)
- \(f'(c)=0\) and \(f\) does not have a relative maximum or relative minimum at \(c\text{.}\)
- \(f\) is not continuous on \([a,b]\) and \(f\) has an absolute maximum at \(c=(a+b)/2\text{.}\)
- \(f\) attains an absolute maximum on \([a,b]\) but does not attain an absolute minimum on \([a,b]\text{.}\)
We first consider a Theorem that shows that if a differentiable function had zero net change on an interval, then somewhere it has a horizontal tangent.
Theorem 1.4.3 Rolle's Theorem
If
- \(f\) is continuous on a closed interval \([a,b]\text{,}\)
- \(f\) is differentiable on \((a,b)\text{,}\) and
- \(f(a)=f(b)\text{,}\)
then there exists \(c\) in \((a,b)\) such that \(f'(c)=0\text{.}\)
Proof
Since \(f\) is continuous on a closed interval \([a,b]\text{,}\) the Extreme Value Theorem 1.4.2 applies, so \(f\) attains both an absolute maximum and an absolute minimum on \([a,b]\text{.}\) We then split into three cases:
Activity 1.4.2
For each of the following cases, either draw a function that satisfies the conditions or explain why it is impossible.
- \(f\) satisfies each of the assumptions of Rolle's Theorem 1.4.3, \(f\) attains its absolute maximum at \(a\text{,}\) and \(f\) attains its absolute minimum at \(a\text{.}\)
- \(f\) satisfies each of the assumptions of Rolle's Theorem 1.4.3 and \(f\) does not attain its absolute maximum at \(a\text{.}\)
- \(f\) satisfies each of the assumptions of Rolle's Theorem 1.4.3 and \(f\) does not attain its absolute minimum at \(a\text{.}\)
Activity 1.4.3
Consider the function \(f(x)=x^5+x+3\text{.}\)
- Use the Intermediate Value Theorem 1.2.1 to show \(f\) has at least one root.
- Use Rolle's Theorem 1.4.3 to show that if \(f\) has more than one root, then \(f'\) has at least one root.
- Show that \(f'(x) \ge 1\text{.}\)
- How many roots does \(f\) have?
Now we turn Rolle's Theorem 1.4.3 on a slant and consider a Theorem that shows somewhere the function has a tangent line with slope the same as its mean rate of change.
Theorem 1.4.4 The Mean Value Theorem
If
- \(f\) is continuous on a closed interval \([a,b]\) and
- \(f\) is differentiable on \((a,b)\text{,}\)
then there exists \(c\) in \((a,b)\) such that
\begin{equation*}
f'(c)=\frac{f(b)-f(a)}{b-a}\text{.}
\end{equation*}
Proof
Let
\begin{equation*}
g(x)=f(x)
-\frac{f(b)-f(a)}{b-a}(x-a)\,
\end{equation*}
for which we can compute
\begin{equation*}
g'(x)=f'(x)
-\frac{f(b)-f(a)}{b-a}\,.
\end{equation*}
Since \(f\) is continuous on \([a,b]\text{,}\) so is \(g\text{.}\) Since \(f\) is differentiable on \((a,b)\text{,}\) so is \(g\text{.}\) We can check directly that \(g(a)=g(b)=f(a)\text{.}\) Thus Rolle's Theorem 1.4.3 applies to \(g\text{,}\) and there is \(c\) in \((a,b)\) that satisfies \(g'(c)=0\text{.}\) Plugging into \(g'(x)\) gives
\begin{equation*}
0=g'(c)=f'(c)
-\frac{f(b)-f(a)}{b-a}
\end{equation*}
so
\begin{equation*}
f'(c)=\frac{f(b)-f(a)}{b-a}\text{.}
\end{equation*}
Activity 1.4.4
- Draw a function \(f\) that satisfies each of the hypotheses of the Mean Value Theorem 1.4.4 and has \(f(a)\not =
f(b)\text{.}\)
- Draw the secant line connecting \((a,f(a))\) and \((b,f(b))\text{.}\) Compute its slope in terms of \(a\text{,}\) \(b\text{,}\) \(f(a)\text{,}\) and \(f(b)\text{.}\)
- Find a point on your graph of \(f\) where the tangent line is parallel to this secant line.
Activity 1.4.5
Suppose we know \(f\) is a differentiable function with \(f(1)=5\) and \(f'(x) \ge 2\text{.}\)
- Sketch several such functions on the interval \([0,4]\)
- What is the greatest possible value for \(f(0)\text{?}\)
- What is the greatest possible value for \(f(4)\text{?}\)
- What is the least possible value for \(f(0)\text{?}\)
- What is the least possible value for \(f(4)\text{?}\)
Justify your answers using the Mean Value Theorem 1.4.4.
Subsection Exercises
1
Consider the function \(f(x)=x^3-4x\) on the interval \([-2,2]\text{.}\)
- Verify that this function satisfies each of the hypotheses of Rolle's Theorem 1.4.3 on this interval.
- Find all numbers \(c\) that satisfy the conclusion of Rolle's Theorem 1.4.3.
2
Consider the function \(f(x)=1/x\) on the interval \([1,3]\text{.}\)
- Verify that this function satisfies each of the hypotheses of the Mean Value Theorem 1.4.4 on this interval.
- Find all numbers \(c\) that satisfy the conclusion of the Mean Value Theorem 1.4.4.
3
If \(f\) is a differentiable function with \(f(0)=2\) and \(f'(x) \le 4\text{,}\) then what is the maximum value \(f(3)\) could be? What is the minimum value \(f(3)\) could be?.