Section 5.4 Applications

The Mean Value Theorem turns derivative information into comparisons of function values. This lets us read off monotonicity, detect local extrema, and prove qualitative facts about derivatives themselves. At the end of the section we include a brief preview of the Fundamental Theorem of Calculus, assuming the usual properties of the definite integral.

Subsection 5.4.1 Monotonicity

Proposition 5.4.1.

Let \(I\) be an open interval, and let \(f \colon I \to \R\) be differentiable on \(I\text{.}\)

The function \(f\) is increasing on \(I\) if and only if \(f'(x) \ge 0\) for every \(x \in I\text{.}\)
The function \(f\) is decreasing on \(I\) if and only if \(f'(x) \le 0\) for every \(x \in I\text{.}\)

Proof.

We prove the increasing case; the decreasing case follows by applying the same argument to \(-f\text{.}\) First suppose that \(f\) is increasing on \(I\text{.}\) Fix \(c \in I\text{.}\) By Carathéodory’s Criterion (Proposition 5.1.4), there exists a function \(\varphi \colon I \to \R\text{,}\) continuous at \(c\text{,}\) such that

\begin{equation*} f(x)-f(c)=\varphi(x)(x-c) \end{equation*}

for all \(x \in I\text{,}\) and \(\varphi(c)=f'(c)\text{.}\) For \(x \neq c\text{,}\) monotonicity of \(f\) gives

\begin{equation*} \varphi(x)=\frac{f(x)-f(c)}{x-c} \ge 0. \end{equation*}

Since \(\varphi\) is continuous at \(c\text{,}\) it follows that \(f'(c)=\varphi(c) \ge 0\text{.}\)

Conversely, suppose that \(f'(x) \ge 0\) for every \(x \in I\text{.}\) Let \(x_1,x_2 \in I\) with \(x_1 \lt x_2\text{.}\) By the Mean Value Theorem (Theorem 5.3.3), applied to the restriction of \(f\) to \([x_1,x_2]\text{,}\) there exists \(c \in (x_1,x_2)\) such that

\begin{equation*} f(x_2)-f(x_1)=f'(c)(x_2-x_1). \end{equation*}

Since \(f'(c) \ge 0\) and \(x_2-x_1 \gt 0\text{,}\) we have \(f(x_2)-f(x_1) \ge 0\text{.}\) Hence \(f(x_1) \le f(x_2)\text{,}\) so \(f\) is increasing on \(I\text{.}\)

Proposition 5.4.2.

Let \(I\) be an open interval, and let \(f \colon I \to \R\) be differentiable on \(I\text{.}\)

If \(f'(x) \gt 0\) for every \(x \in I\text{,}\) then \(f\) is strictly increasing on \(I\text{.}\)
If \(f'(x) \lt 0\) for every \(x \in I\text{,}\) then \(f\) is strictly decreasing on \(I\text{.}\)

Proof.

Again we prove only the increasing case. Let \(x_1,x_2 \in I\) with \(x_1 \lt x_2\text{.}\) By the Mean Value Theorem (Theorem 5.3.3), there exists \(c \in (x_1,x_2)\) such that

\begin{equation*} f(x_2)-f(x_1)=f'(c)(x_2-x_1). \end{equation*}

Since \(f'(c) \gt 0\) and \(x_2-x_1 \gt 0\text{,}\) we get \(f(x_2)-f(x_1) \gt 0\text{.}\) Therefore \(f\) is strictly increasing on \(I\text{.}\)

Example 5.4.3.

The converse of Proposition 5.4.2 is false. The function \(f(x)=x^3\) is strictly increasing on \(\R\text{,}\) but \(f'(0)=0\text{.}\)

The graph of y equals x cubed rises from left to right and has a horizontal tangent at the origin. — Figure 5.4.4. The function \(x^3\) is strictly increasing even though its tangent is horizontal at \(0\text{.}\)

Corollary 5.4.5.

Let \(I\) be an open interval, and let \(f \colon I \to \R\) be differentiable on \(I\text{.}\) If \(f'(x)=0\) for every \(x \in I\text{,}\) then \(f\) is constant on \(I\text{.}\)

Proof.

By Proposition 5.4.1, \(f\) is both increasing and decreasing on \(I\text{.}\) Hence \(f\) is constant on \(I\text{.}\)

The fact that the domain is an interval is essential here. For example, define \(f \colon (-1,0)\cup(0,1) \to \R\) by

\begin{equation*} f(x)=\begin{cases} -1 & \text{if } x \lt 0,\\ 1 & \text{if } x \gt 0. \end{cases} \end{equation*}

Then \(f'(x)=0\) for every point of its domain, but \(f\) is not constant. What remains true in general is that \(f\) is constant on each interval component of its domain.

Subsection 5.4.2 Derivative Tests for Extrema

The sign of the derivative often reveals where extrema occur. The first derivative test uses only the sign pattern of \(f'\text{.}\) The second derivative test packages the same idea into a convenient condition at a single point.

Proposition 5.4.6. First Derivative Test.

Let \(c \in (a,b)\text{,}\) and let \(f \colon (a,b) \to \R\) be continuous on \((a,b)\) and differentiable on \((a,b)\setminus\{c\}\text{.}\)

If \(f'(x) \ge 0\) on \((a,c)\) and \(f'(x) \le 0\) on \((c,b)\text{,}\) then \(c\) is a maximum point of \(f\) on \((a,b)\text{.}\)
If \(f'(x) \le 0\) on \((a,c)\) and \(f'(x) \ge 0\) on \((c,b)\text{,}\) then \(c\) is a minimum point of \(f\) on \((a,b)\text{.}\)

If the inequalities are strict, then the corresponding extremum is strict.

Proof.

We prove the first statement. Let \(x \in (a,c)\text{.}\) The function \(f\) is continuous on \([x,c]\) and differentiable on \((x,c)\text{,}\) so the Mean Value Theorem (Theorem 5.3.3) gives a point \(u_x \in (x,c)\) such that

\begin{equation*} f(c)-f(x)=f'(u_x)(c-x). \end{equation*}

Since \(f'(u_x) \ge 0\) and \(c-x \gt 0\text{,}\) we obtain \(f(c)-f(x) \ge 0\text{,}\) that is, \(f(x) \le f(c)\text{.}\)

Similarly, if \(y \in (c,b)\text{,}\) then applying the Mean Value Theorem to \([c,y]\) gives \(v_y \in (c,y)\) such that

\begin{equation*} f(y)-f(c)=f'(v_y)(y-c). \end{equation*}

Since \(f'(v_y) \le 0\) and \(y-c \gt 0\text{,}\) we get \(f(y) \le f(c)\text{.}\) Therefore \(f(x) \le f(c)\) for every \(x \in (a,b)\text{,}\) so \(c\) is a maximum point of \(f\text{.}\) The strict case is identical, with \(\ge\) replaced by \(\gt\) and \(\le\) replaced by \(\lt\text{.}\) The second statement follows by applying the first to \(-f\text{.}\)

Proposition 5.4.7. Second Derivative Test.

Let \(f\) be defined on a neighborhood of \(x_0\text{,}\) and suppose that \(f\) is twice differentiable at \(x_0\) with \(f'(x_0)=0\text{.}\)

If \(f''(x_0) \gt 0\text{,}\) then \(x_0\) is a strict local minimum of \(f\text{.}\)
If \(f''(x_0) \lt 0\text{,}\) then \(x_0\) is a strict local maximum of \(f\text{.}\)

Proof.

We prove the first statement. Apply Carathéodory’s Criterion (Proposition 5.1.4) to \(f'\) at \(x_0\text{.}\) Then there exists a function \(\psi\text{,}\) continuous at \(x_0\text{,}\) such that

\begin{equation*} f'(x)-f'(x_0)=\psi(x)(x-x_0) \end{equation*}

for all \(x\) near \(x_0\text{,}\) and \(\psi(x_0)=f''(x_0) \gt 0\text{.}\) Since \(f'(x_0)=0\text{,}\) this becomes

\begin{equation*} f'(x)=\psi(x)(x-x_0). \end{equation*}

Because \(\psi\) is continuous at \(x_0\) and \(\psi(x_0) \gt 0\text{,}\) there exists \(\delta \gt 0\) such that \(\psi(x) \gt 0\) whenever \(|x-x_0| \lt \delta\text{.}\) Hence \(f'(x) \lt 0\) on \((x_0-\delta,x_0)\) and \(f'(x) \gt 0\) on \((x_0,x_0+\delta)\text{.}\) By the First Derivative Test (Proposition 5.4.6), \(x_0\) is a strict local minimum of \(f\text{.}\) The second statement follows by applying the first to \(-f\text{.}\)

Subsection 5.4.3 Darboux’s Theorem and a Preview of the FTC

Theorem 5.4.8. Darboux’s Theorem.

Let \(f \colon [a,b] \to \R\) be differentiable on \([a,b]\text{.}\) If \(\lambda\) lies between \(f'(a)\) and \(f'(b)\text{,}\) then there exists \(c \in (a,b)\) such that

\begin{equation*} f'(c)=\lambda. \end{equation*}

Proof.

Replacing \(f\) by \(x \mapsto f(x)-\lambda x\text{,}\) it suffices to prove the theorem in the case \(\lambda=0\text{.}\) Replacing \(f\) by \(-f\) if necessary, we may further assume that \(f'(a) \lt 0 \lt f'(b)\text{.}\)

Since \(f\) is differentiable at \(a\text{,}\) Carathéodory’s Criterion (Proposition 5.1.4) gives a function \(\varphi_a\text{,}\) continuous at \(a\text{,}\) such that

\begin{equation*} f(x)-f(a)=\varphi_a(x)(x-a) \end{equation*}

for all \(x \in [a,b]\text{,}\) with \(\varphi_a(a)=f'(a) \lt 0\text{.}\) Hence \(\varphi_a(x) \lt 0\) for all \(x \gt a\) sufficiently close to \(a\text{.}\) Since also \(x-a \gt 0\text{,}\) it follows that \(f(x)-f(a) \lt 0\) for such \(x\text{.}\) Therefore \(a\) is not a minimum point of \(f\) on \([a,b]\text{.}\)

A similar argument at \(b\) shows that \(b\) is not a minimum point either. Indeed, Carathéodory’s Criterion gives a function \(\varphi_b\text{,}\) continuous at \(b\text{,}\) such that

\begin{equation*} f(x)-f(b)=\varphi_b(x)(x-b) \end{equation*}

for all \(x \in [a,b]\text{,}\) with \(\varphi_b(b)=f'(b) \gt 0\text{.}\) Thus \(\varphi_b(x) \gt 0\) for all \(x \lt b\) sufficiently close to \(b\text{.}\) Since then \(x-b \lt 0\text{,}\) we obtain \(f(x)-f(b) \lt 0\text{.}\) Hence \(b\) is not a minimum point of \(f\text{.}\)

By the Extreme Value Theorem (Theorem 4.4.4), \(f\) attains its minimum value at some point \(c \in [a,b]\text{.}\) Thus, we must have \(c \in (a,b)\text{.}\) The Critical Point Theorem (Proposition 5.3.1) then gives \(f'(c)=0\text{,}\) as required.

Darboux’s theorem shows that derivatives have the intermediate value property, even though derivatives need not be continuous. We now record one further application of the earlier results. This final part depends on the usual properties of the definite integral, which we have not yet developed systematically, so it should be viewed as a preview.

Proposition 5.4.9. Integral Mean Value Theorem.

Assume the basic properties of the definite integral. If \(f \colon [a,b] \to \R\) is continuous, then there exists \(c \in [a,b]\) such that

\begin{equation*} \int_a^b f(x)\,dx = f(c)(b-a). \end{equation*}

Proof.

By the Extreme Value Theorem (Theorem 4.4.4), \(f\) attains an absolute minimum \(m\) and an absolute maximum \(M\) on \([a,b]\text{.}\) Hence

\begin{equation*} m \le f(x) \le M \quad \text{for all } x \in [a,b]. \end{equation*}

Integrating gives

\begin{equation*} m(b-a) \le \int_a^b f(x)\,dx \le M(b-a). \end{equation*}

Since \(b-a \gt 0\text{,}\)

\begin{equation*} m \le \frac{1}{b-a}\int_a^b f(x)\,dx \le M. \end{equation*}

Because \(f\) is continuous on \([a,b]\text{,}\) the Intermediate Value Theorem (Theorem 4.4.8) yields some \(c \in [a,b]\) such that

\begin{equation*} f(c)=\frac{1}{b-a}\int_a^b f(x)\,dx. \end{equation*}

Multiplying by \(b-a\) gives the result.

Theorem 5.4.10. Fundamental Theorem of Calculus (Special Case).

Assume the basic properties of the definite integral. If \(f \colon [a,b] \to \R\) is continuous and

\begin{equation*} F(x)=\int_a^x f(t)\,dt, \end{equation*}

then \(F\) is differentiable on \([a,b]\) and

\begin{equation*} F'(x)=f(x) \end{equation*}

for every \(x \in [a,b]\text{.}\)

Proof.

Fix \(x \in [a,b]\text{.}\) Let \(h \neq 0\) be small enough that \(x+h \in [a,b]\text{.}\) Then

\begin{equation*} F(x+h)-F(x)=\int_x^{x+h} f(t)\,dt. \end{equation*}

By the Integral Mean Value Theorem (Proposition 5.4.9), there exists a point \(c_h\) between \(x\) and \(x+h\) such that

\begin{equation*} \int_x^{x+h} f(t)\,dt = f(c_h)h. \end{equation*}

Therefore

\begin{equation*} \frac{F(x+h)-F(x)}{h}=f(c_h). \end{equation*}

Since \(c_h\) lies between \(x\) and \(x+h\text{,}\) we have \(c_h \to x\) as \(h \to 0\text{.}\) By continuity of \(f\text{,}\) \(f(c_h) \to f(x)\text{.}\) Hence

\begin{equation*} \lim_{h\to 0}\frac{F(x+h)-F(x)}{h}=f(x), \end{equation*}

so \(F'(x)=f(x)\text{.}\)

This theorem is only a preview. In the integration chapter we prove the more general second form of the Fundamental Theorem of Calculus (Theorem 7.3.1), where the integrand is assumed only to be Riemann integrable and the conclusion is differentiability at points of continuity of the integrand.

Corollary 5.4.11.

Assume the basic properties of the definite integral. If \(f \colon [a,b] \to \R\) is continuous and \(F\) is any antiderivative of \(f\) on \([a,b]\text{,}\) then

\begin{equation*} \int_a^b f(x)\,dx = F(b)-F(a). \end{equation*}

Proof.

Let \(F_a(x)=\int_a^x f(t)\,dt\text{.}\) By the Fundamental Theorem of Calculus (Special Case) (Theorem 5.4.10), both \(F\) and \(F_a\) are antiderivatives of \(f\) on \([a,b]\text{.}\) Hence \((F-F_a)'(x)=0\) for every \(x \in (a,b)\text{.}\) By Corollary 5.4.5, the function \(F-F_a\) is constant on \((a,b)\text{.}\) Since both \(F_a\) and \(F\) are continuous on \([a,b]\text{,}\) it follows that \(F-F_a\) is constant on all of \([a,b]\text{.}\)

Evaluating at \(x=a\text{,}\) we obtain

\begin{equation*} F(a)-F_a(a)=F(a), \end{equation*}

because \(F_a(a)=\int_a^a f(t)\,dt=0\text{.}\) Therefore \(F(x)-F_a(x)=F(a)\) for every \(x \in [a,b]\text{.}\) Setting \(x=b\) gives

\begin{equation*} F(b)-\int_a^b f(t)\,dt = F(a), \end{equation*}

and rearranging yields

\begin{equation*} \int_a^b f(x)\,dx = F(b)-F(a). \end{equation*}

Likewise, the integration chapter proves the more general first form of the Fundamental Theorem of Calculus (Theorem 7.3.5) and the corresponding Newton-Leibniz formula (Corollary 7.3.6).

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