Section 15.3 The Proof of the First Derivative Test
We stated
Theorem 9.5.4 in
Chapter 9 without stating the conditions that make it true, because at the time our attention was primarily on using it.
Since the focus of our attention here is on what makes the First Derivative Test true we will restate it to reflect our new, and deeper, understanding. Notice that the conclusion is the same, only the conditions have changed.
Theorem 15.3.1. First Derivative Test.
Suppose \(f(x)\) is continuous on the interval \([\alpha,\beta]\text{,}\) and differentiable on the interval \((\alpha,\beta)\text{.}\) Suppose further that both \(a\) and \(b\) are in the interval \([\alpha,\beta]\) and \(b\gt a\text{.}\)
If \(f^\prime(x)\gt 0\) on the interval \((\alpha,\beta)\) then \(f(b)\gt f(a)\text{.}\) (That is, the function is increasing on \([\alpha,\beta]\text{.}\))
If \(f^\prime(x)\lt 0\) on the interval \((\alpha,\beta)\) then \(f(b)\lt f(a)\text{.}\) (That is, the function is decreasing on \([\alpha,\beta]\text{.}\))
Proof of part (a) of Theorem 15.3.1.
We want to use the Mean Value Theorem on the interval \([a,b]\) so we begin by verifying that the conditions of the Mean Value Theorem are satisfied on that interval. Observe that \([a,b]\) is a subinterval of \([\alpha,\beta]\) so \(f(x)\) is continuous on \([a,b]\) and differentiable on \((a,b)\text{.}\) By the Mean Value Theorem there is a number, \(c\text{,}\) in the interval \((a,b)\) such that
\begin{align*}
\frac{f(b)-f(a)}{b-a} \amp= f^\prime(c)\\
f(b)-f(a) \amp= f^\prime(c)(b-a).
\end{align*}
Since both \(b-a\gt 0\) and \(f^\prime(c)\gt 0\) are positive, \(f(b)-f(a)\) must be positive as well. Therefore
\begin{equation*}
f(b)-f(a) \gt 0 \text{ or }
f(b)\gt f(a).
\end{equation*}
Problem 15.3.2.
(a)
By modifying our proof of part (a) of the First Derivative Test as needed.
(b)
Let \(g(x)=-f(x)\) and apply part (a) of this problem. (Don’t forget to show first that \(g(x)\) satisfies the conditions of the Mean Value Theorem.)
While we are in this frame of mind, we’ll take a moment to notice that we can use the Mean Value Theorem to prove, rigorously, something that we have alluded to a few times but have never addressed directly. It is clear from our differentiation rules that if two functions differ by a constant, then they have the same derivative. We’ve mentioned that the converse is true, namely if two functions have the same derivative on an interval then they must differ by a constant. This can be proved in a manner similar to the proof above.
Problem 15.3.3.
(a)
Suppose \(f^\prime(x)=0\) on the interval \((\alpha,\beta)\) and that \(a\) and \(b\) are two points in that interval. Use an argument similar to the proof of the first derivative test to show that \(f(a)=f(b)\text{.}\)
(b)
Explain how the result of part (a) says that \(f(x)\) must be constant on \((\alpha, \beta)\text{.}\)
(c)
Show that if \(f^\prime(x)=g^\prime(x)\) on the interval \((\alpha,\beta)\text{,}\) then
\begin{equation*}
f(x)=g(x)+c
\end{equation*}
for some constant \(c\text{.}\)
Hint.
Consider the function \(F(x)=f(x)-g(x)\text{.}\)
(d)
What can be said if \(f^\prime(x)=g^\prime(x)\) for all \(x\) in some set \(S\) which is not an interval?
Problem 15.3.4.
\(f(x)\) and \(g(x)\) are differentiable on an open interval containing \(a\text{.}\)
\(f(a)=g(a)=0\text{.}\)
\(\limit{x}{a}{f^\prime(x)} =L \) and \(\limit{x}{a}{g^\prime(x)}
\ne 0 \text{.}\)
(a)
Under these conditions explain why
\begin{equation*}
\limit{x}{a}{\frac{f(x)}{g(x)}}=\limit{x}{a}{\frac{f(x)-f(a)}{g(x)-g(a)}}
\end{equation*}
(b)
Apply the Mean Value Theorem to the numerator and denominator part (a) and then prove that
\begin{equation*}
\limit{x}{a}{\frac{f(x)}{g(x)}}\lhopeq\frac{L}{M}\text{.}
\end{equation*}
Hint.
Use the fact that in the statement of
Theorem 15.2.4 \(c\) is between
\(a\) and
\(b\text{.}\)
