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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 $f$.

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.

$\Box$

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.

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