[Calculus] VI. Applications of Differentiation

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Maximum and Minimum Values

Definition 6.1.1.  Let \( c \) be a number in the domain \( D \) of a function \( f \). Then \( f(c) \) is the

·  absolute maximum value of \( f \) on \( D \) if \( f(c) \ge f(x) \) for all \( x \) in \( D \).

·  absolute minimum value of \( f \) on \( D \) if \( f(c) \le f(x) \) for all \( x \) in \( D \).

Remark 6.1.2.  An absolute maximum or minimum is sometimes called a global maximum or minimum. The maximum and minimum values of \( f \) are called extreme values of \( f \).

Definition 6.1.3.  The number \( f(c) \) is a

·  local maximum value of \( f \) if \( f(c) \ge f(x) \) when \( x \) is near \( c \).

·  local minimum value of \( f \) if \( f(c) \le f(x) \) when \( x \) is near \( c \).

Remark 6.1.4.  If we say that something is true near \( c \), we mean that it is true on some open interval containing \( c \).

Figure 1

Theorem 6.1.5. (The Extreme Value Theorem)  If \( f \) is continuous on a closed interval \( \left[a,\ b\right] \), then \( f \) attains an absolute maximum value \( f(c) \) and an absolute minimum value \( f(d) \) at some numbers \( c \) and \( d \) in \( \left[a,\ b \right] \).

Figure 2

Theorem 6.1.6. (Fermat's Theorem)  If \( f \) has a local maximum or minimum at \( c \), and if \( f'(c) \) exists, then \( f'(c) = 0 \).

Figure 3

Proof.  Suppose, for the sake of definiteness, that \( f \) has a local maximum at \( c \). Then, according to Definition 6.1.3, \( f(c) \ge f(x) \) if \( x \) is sufficiently close to \( c \). This implies that if \( h \) is sufficiently close to \( 0 \), with \( h \) being positive or negative, then

\( \displaystyle f(c) \ge f(c + h) \)

and therefore

\( \displaystyle f(c + h) - f(c) \le 0 \qquad \cdots \qquad (\text{i}) \)

We can divide both sides of an inequality by a positive number. Thus, if \( h > 0 \) and \( h \) is sufficiently small, we have

\( \displaystyle \frac{f(c + h) - f(c)}{h} \le 0 \)

Taking the right-hand limit of both sides of this inequality (using Theorem 1.2.2), we get

\( \displaystyle \lim_{h \to 0^{+}}\!\frac{f(c + h) - f(c)}{h} \le \lim_{h \to 0^{+}}\!0 = 0 \)

But since \( f'(c) \) exists, we have

\( \displaystyle f'(c) = \lim_{h \to 0}\frac{f(c + h) - f(c)}{h} = \lim_{h \to 0^{+}}\!\frac{f(c + h) - f(c)}{h} \)

and so we have shown that \( f'(c) \le 0 \).

If \( h < 0 \), then the direction of the inequality \( (\text{i}) \) is reversed when we divide by \( h \):

\( \displaystyle \frac{f(c + h) - f(c)}{h} \ge 0 \)

So, taking the left-hand limit, we have

\( \displaystyle f'(c) = \lim_{h \to 0}\frac{f(c + h) - f(c)}{h} = \lim_{h \to 0^{-}}\!\frac{f(c + h) - f(c)}{h} \ge 0 \)

We have shown that \( f'(c) \ge 0 \) and also that \( f'(c) \le 0 \). Since both of these inequalities must be true, the only possibility is that \( f'(c) = 0 \).

\( \Box \)

Remark 6.1.7.  You must be careful that the converse of Theorem 6.1.6 is false in general. In other words, even when \( f'(c) = 0 \) there need not be a maximum or minimum at \( c \). Furthermore, there may be an extreme value even when \( f'(c) \) does not exist.

Definition 6.1.8.  A critical number of a function \( f \) is a number \( c \) in the domain of \( f \) such that either \( f'(c) = 0 \) or \( f'(c) \) does not exist.

Remark 6.1.9.  If \( f \) has a local maximum or minimum at \( c \), then \( c \) is a critical number of \( f \).

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The Mean Value Theorem

Theorem 6.2.1. (Rolle’s Theorem)  Let \( f \) be a function that satisfies the following three hypotheses:

1.  \( f \) is continuous on the closed interval \( \left[a,\ b \right] \).

2.  \( f \) is differentiable on the open interval \( \left(a,\ b \right) \).

3.  \( f(a) = f(b) \)

Then there is a number \( c \) in \( \left(a,\ b \right) \) such that \( f'(c) = 0 \).

Figure 4

Proof.  There are three cases:

Case I.  \( f(x) = k \), a constant

Then \( f'(x) = 0 \), so the number \( c \) can be taken to be any number in \( \left(a,\ b \right) \).

Case II.  \( f(x) > f(a) \) for some \( x \) in \( \left(a,\ b \right) \)

By Theorem 6.1.5 (which we can apply by hypothesis 1), \( f \) has a maximum value somewhere in \( \left[a,\ b \right] \). Since \( f(a) = f(b) \), it must attain this maximum value at a number \( c \) in the open interval \( \left(a,\ b \right) \). Then \( f \) has a local maximum at \( c \) and, by hypothesis 2, \( f \) is differentiable at \( c \). Therefore \( f'(c) = 0 \) by Theorem 6.1.6.

Case III.  \( f(x) < f(a) \) for some \( x \) in \( \left(a,\ b \right) \)

By Theorem 6.1.5, \( f \) has a minimum value in \( \left[a,\ b \right] \) and, since \( f(a) = f(b) \), it attains this minimum value at a number \( c \) in \( \left(a,\ b \right) \). Again \( f'(c) = 0 \) by Theorem 6.1.6.

\( \Box \)

Theorem 6.2.2. (The Mean Value Theorem)  Let \( f \) be a function that satisfies the following hypotheses:

1.  \( f \) is continuous on the closed interval \( \left[a,\ b \right] \).

2.  \( f \) is differentiable on the open interval \( \left(a,\ b \right) \).

Then there is a number \( c \) in \( \left(a,\ b \right) \) such that

\( \displaystyle f'(c) = \frac{f(b) - f(a)}{b - a} \qquad \cdots \qquad \text{(ii)} \)

or, equivalently,

\( \displaystyle f(b) - f(a) = f'(c)(b - a) \)

Figure 5

Proof.  The slope of the secant line \( AB \) is

\( \displaystyle m_{AB} = \frac{f(b) - f(a)}{b - a} \qquad \cdots \qquad \text{(iii)} \)

which is the same expression as on the right side of Equation \( \text{(ii)} \).

Now we apply Theorem 6.2.1 to a new function \( h \) defined as the difference between \( f \) and the function whose graph is the secant line \( AB \). Using Equation \( \text{(iii)} \) and the point-slope equation of a line, we see that the equation of the line \( AB \) can be written as

\( \displaystyle y - f(a) = \frac{f(b) - f(a)}{b - a}(x - a) \)

or as

\( \displaystyle y = f(a) + \frac{f(b) - f(a)}{b - a}(x - a) \)

So, the equation of \( h(x) \) is

\( \displaystyle h(x) = f(x) - f(a) - \frac{f(b) - f(a)}{b - a}(x - a) \qquad \cdots \qquad \text{(iv)} \)

First we must verify that \( h \) satisfies the three hypotheses of Theorem 6.2.1.

1.  The function \( h \) is continuous on \( \left[a,\ b \right] \) because it is the sum of \( f \) and a first-degree polynomial, both of which are continuous.

2.  The function \( h \) is differentiable on \( \left(a,\ b \right) \) because both \( f \) and the first-degree polynomial are differentiable. In fact, we can compute \( h' \) directly from Equation \( \text{(iv)} \):

\( \displaystyle h'(x) = f'(x) - \frac{f(b) - f(a)}{b - a} \)

3.

\( \displaystyle \begin{align} h(a) &= f(a) - f(a) - \frac{f(b) - f(a)}{b - a}(a - a) = 0 \\[14pt] h(b) &= f(b) - f(a) - \frac{f(b) - f(a)}{b - a}(b - a) \\[7pt] &= f(b) - f(a) - \left[ f(b) - f(a) \right] = 0 \end{align} \)

Therefore \( h(a) = h(b) \).

Since \( h \) satisfies the hypotheses of Theorem 6.2.1, that theorem says there is a number \( c \) in \( \left(a,\ b \right) \) such that \( h'(c) = 0 \). Therefore

\( \displaystyle 0 = h'(c) = f'(c) - \frac{f(b) - f(a)}{b - a} \)

and so

\( \displaystyle f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( \Box \)

Theorem 6.2.3.  If \( f'(x) = 0 \) for all \( x \) in an interval \( \left(a,\ b\right) \), then \( f \) is constant on \( \left(a,\ b\right) \).

Proof.  Let \( x_{1} \) and \( x_{2} \) be any two numbers in \( \left(a,\ b\right) \) with \( x_{1} < x_{2} \). Since \( f \) is differentiable on \( \left(a,\ b\right) \), it must be differentiable on \( \left(x_{1},\ x_{2}\right) \) and continuous on \( \left[x_{1},\ x_{2}\right] \). By applying Theorem 6.2.2. to \( f \) on the interval \( \left[x_{1},\ x_{2}\right] \), we get a number \( c \) such that \( x_{1} < c < x_{2} \) and

\( \displaystyle f(x_{2}) - f(x_{1}) = f'(c)(x_{2} - x_{1}) \qquad \cdots \qquad \text{(v)} \)

Since \( f'(x) = 0 \) for all \( x \), we have \( f'(c) = 0 \), and so Equation \( \text{(v)} \) becomes

\( \displaystyle f(x_{2}) - f(x_{1}) = 0 \qquad \text{or} \qquad f(x_{2}) = f(x_{1}) \)

Therefore \( f \) has the same value at any two numbers \( x_{1} \) and \( x_{2} \) in \( \left(a,\ b\right) \). This means that \( f \) is constant on \( \left(a,\ b\right) \).

\( \Box \)

Corollary 6.2.4.  If \( f'(x) = g'(x) \) for all \( x \) in an interval \( \left(a,\ b\right) \), then \( f - g \) is constant on \( \left(a,\ b\right) \); that is, \( f(x) = g(x) + c \) where \( c \) is a constant.

Proof.  Let \( F(x) = f(x) - g(x) \). Then

\( \displaystyle F'(x) = f'(x) - g'(x) = 0 \)

for all \( x \) in \( \left(a,\ b\right) \). Thus, by Theorem 6.2.3, \( F \) is constant; that is, \( f - g \) is constant.

\( \Box \)

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The Shape of a Graph

Theorem 6.3.1.

(a)  If \( f'(x) > 0 \) on an interval, then \( f \) is increasing on that interval.

(b)  If \( f'(x) < 0 \) on an interval, then \( f \) is decreasing on that interval.

Proof.  (a) Let \( x_{1} \) and \( x_{2} \) be any two numbers in the interval with \( x_{1} < x_{2} \). According to the definition of an increasing function, we have to show that \( f(x_{1}) < f(x_{2}) \).

Because we are given that \( f'(x) > 0 \), we know that \( f \) is differentiable on \( \left[x_{1},\ x_{2} \right] \). So, by Theorem 6.2.2, there is a number \( c \) between \( x_{1} \) and \( x_{2} \) such that

\( \displaystyle f(x_{2}) - f(x_{1}) = f'(c)(x_{2} - x_{1}) \qquad \cdots \qquad \text{(vi)} \)

Now \( f'(c) > 0 \) by assumption and \( x_{2} - x_{1} > 0 \) because \( x_{1} < x_{2} \). Thus the right side of Equation \( \text{(vi)} \) is positive, and so

\( \displaystyle f(x_{2}) - f(x_{1}) > 0 \qquad \text{or} \qquad f(x_{1}) < f(x_{2}) \)

This shows that \( f \) is increasing.

\( \Box \)

Theorem 6.3.2.  Suppose that \( c \) is a critical number of a continuous function \( f \).

(a)  If \( f' \) changes from positive to negative at \( c \), then \( f \) has a local maximum at \( c \).

(b)  If \( f' \) changes from negative to positive at \( c \), then \( f \) has a local minimum at \( c \).

(c)  If \( f' \) is positive to the left and right of \( c \), or negative to the left and right of \( c \), then \( f \) has no local maximum or minimum at \( c \).

Figure 6

Remark 6.3.3.  Theorem 6.3.2 is a consequence of Theorem 6.3.1. In part (a), for instance, since the sign of \( f'(x) \) changes from positive to negative at \( c \), \( f \) is increasing to the left of \( c \) and decreasing to the right of \( c \). It follows that \( f \) has a local maximum at \( c \).

Definition 6.3.4.  If the graph of \( f \) lies above all of its tangents on an interval \( I \), then it is called concave upward on \( I \). If the graph of \( f \) lies below all of its tangents on \( I \), it is called concave downward on \( I \).

Figure 7 : Concave Upward

Figure 8 : Concave Downward

Definition 6.3.5.  A point \( P \) on a curve \( y = f(x) \) is called an inflection point if \( f \) is continuous there and the curve changes from concave upward to concave downward or from concave downward to concave upward at \( P \).

Theorem 6.3.6.

(a)  If \( f''(x) > 0 \) for all \( x \) in \( I \), then the graph of \( f \) is concave upward on \( I \).

(b)  If \( f''(x) < 0 \) for all \( x \) in \( I \), then the graph of \( f \) is concave downward on \( I \).

Proof.  (a) Let \( a \) be any number in \( I \). We must show that the curve \( y = f(x) \) lies above the tangent line at the point \( (a, f(a)) \). The equation of this tangent is

\( \displaystyle y = f(a) + f'(a)(x - a) \)

So we must show that

\( \displaystyle f(x) > f(a) + f'(a)(x - a) \)

whenever \( x \in I \ (x \neq a) \).

First let us take the case where \( x > a \). Applying Theorem 6.2.2 to \( f \) on the interval \( \left[a,\ x \right] \), we get a number \( c \), with \( a < c < x \), such that

\( \displaystyle f(x) - f(a) = f'(c)(x - a) \qquad \cdots \qquad \text{(vii)} \)

Since \( f'' > 0 \) on \( I \), we know from Theorem 6.3.1 that \( f' \) is increasing on \( I \). Thus, since \( a < c \), we have

\( \displaystyle f'(a) < f'(c) \)

and so, multiplying this inequality by the positive number \( x - a \), we get

\( \displaystyle f'(a)(x - a) < f'(c)(x - a) \)

Now we add \( f(a) \) to both sides of this inequality:

\( \displaystyle f(a) + f'(a)(x - a) < f(a) + f'(c)(x - a) \)

But from Equation \( \text{(vii)} \) we have \( f(x) = f(a) + f'(c)(x - a) \). So this inequality becomes

\( \displaystyle f(x) > f(a) + f'(a)(x - a) \)

which is what we wanted to prove.

\( \Box \)

Theorem 6.3.7.  Suppose \( f'' \) is continuous near \( c \).

(a)  If \( f'(c) = 0 \) and \( f''(c) > 0 \), then \( f \) has a local minimum at \( c \).

(b)  If \( f'(c) = 0 \) and \( f''(c) < 0 \), then \( f \) has a local maximum at \( c \).

Remark 6.3.8.  Theorem 6.3.7 is a consequence of Theorem 6.3.6 and serves as an alternative to Theorem 6.3.2.

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The body text and the figures were excerpted from James Stewart, Calculus 8E.