Applications of Differentiation

Maximum and Minimum Values

Some of the most important applications of differential calculus are optimization problems, in which we are required to find the optimal (best) way of doing something.

Absolute and Local Extreme Values

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 $f$.

The Extreme Value Theorem is illustrated in Figure 8.

Critical Numbers and the Closed Interval Method

PROOF: Suppose that $f$ has a local maximum at $c$. Then accoding to the definition of a local maximum $f(c) \geq f(x)$ if $x$ is sufficiently close to $c$. This implies that if $h$ is sufficiently close to $0$, with $h$ being positive or negative, then:

$$ f(c) \geq f(c + h) $$

and therefore

$$ f(c + h) - f(c) \leq 0 $$

We can divide both sides of an inequality by a positive number $h$, then

$$ \frac{f(c + h) - f(c)}{h} \leq 0 $$

Taking the right-hand limit of both sides we get

$$ \lim_{h \rightarrow 0^+} \frac{f(c + h) - f(c)}{h} \leq \lim_{h \rightarrow 0^+} = 0 $$

Since $f’(0)$ exists, we have

$$ \lim_{h \rightarrow 0} \frac{f(c + h) - f(c)}{h} = \lim_{h \rightarrow 0^+} \frac{f(c + h) - f(c)}{h} $$

and so we have shown that $f’(c) \leq 0$. If $h < 0$, then

$$ \frac{f(c + h) - f(c)}{h} \geq 0 $$

So taking the left-hand limit, we have

$$ \lim_{h \rightarrow 0} \frac{f(c + h) - f(c)}{h} = \lim_{h \rightarrow 0^-} \frac{f(c + h) - f(c)}{h} \geq 0 $$

We have shown that $f’(c) \geq 0$ and $f’(c) \leq 0$ so the only possibility is that $f’(c) = 0$.

We can’t expect to locate extreme values simply by setting $f’(x) = 0$ and solving for $x$. The fact that $f’(x) = 0$ simply means that the curve $f$ has a horizontal tangent at $(x, f(x))$. In other words, the converse of Fermat’s Theorem is false in general. However, Fermat’s Theorem does suggest that we should at least start looking for extreme values of $f$ at the number $c$ where $f’(c) = 0$ or where $f’(c)$ does not exist.

In terms of critical numbers, Fermat’s Theorem can be rephared as follows:

The Mean Value Theorem

Rolle’s Theorem

PROOF: There are three cases:

Case 1 $f(x) = k$, a constant. Then $f’(x) = 0$, so the number $c$ can be taken to be any numer in $(a, b)$.

Case 2 $f(x) > f(a)$ for some $x \in (a, b)$. Then, by the Extreme Value Theorem (which we can apply by hypothesis 1), $f$ has a maximum value somewhere in $[a, b]$. Since $f(a) = f(b)$ it must attain this maximum value at a number $c$ in the open interval $(a, b)$. Then $f$ has a local maximum at $c$, and by hypothesis 2, $f$ is differentiable at $c$. Therefore $f’(c) = 0$ by Fermat’s Theorem.

Case 3 $f(x) < f(a)$ for some $x \in (a, b)$. Then, by the Extreme Value Theorem (which we can apply by hypothesis 1), $f$ has a minimum value somewhere in $[a, b]$. Since $f(a) = f(b)$ it must attain this minimum value at a number $c$ in the open interval $(a, b)$. Then $f$ has a local minimum at $c$, and by hypothesis 2, $f$ is differentiable at $c$. Therefore $f’(c) = 0$ by Fermat’s Theorem.

Figure 1 shows the graphs of four such functions.

The Mean Value Theorem

The Mean Value Theorem says that there is at least one point $P(c, f(c))$ on the graph where the slope of the tangent line is the same as the slope of the secant line $A, B$ (see Figures 3 and 4).

PROOF We see the the equation of the line $AB$ can be written as:

$$ y - f(a) = m_{AB} (x - a) $$

where $m_{AB} = \frac{f(b) - f(a)}{b - a}$, thus

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

So, as we can see on Figure 5:

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

First, we must verify that $h$ satisfies the three hypotheses of Rolle’s Theorem:

1. The function $h$ is continuous on $[a, b]$ because it is the sum of $f$ and a first-degree polynomial both of which are continuous.
2. The function $h$ is differentiable because both $f$ and the first degree polynomial are differentiable. Thus

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

3. We must show that $h(a) = h(b)$

$$ h(a) = f(a) - f(a) - \frac{f(b) - f(a)}{b - a}(a - a) = 0 $$$$ h(b) = f(b) - f(a) - \frac{f(b) - f(a)}{b - a}(b - a) $$$$ = f(b) - f(a) - f(b) + f(a) = 0 $$

Which means $h(a) = h(b) = 0$.

Since $h$ satisfies all the hypotheses of Rolle’s Theorem, there exists a number $c$ in $(a, b)$ such that $h’(c) = 0$. Therefore

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

So

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

In general, the Mean Value Theorem can be interpreted as saying that there is a number at which the instantaneous rate of change is equal to the average rate of change over an interval.

The Mean Value Theorem can be used to establish some of the basic facts of differential calculus.

PROOF: Let $x_1$ and $x_2$ be any two numbers in $(a, b)$ with $x_1 < x_2$. Since $f$ is differentiable on $(a, b)$ it must be differentiable on $(x_1, x_2)$ and continuous on $[x_1, x_2]$. By applying the Mean Value Theorem to $f$ on the interval $[x_1, x_2]$, we get a number $c$ such that $x_1 < c < x_2$ and

$$ f(x_2) - f(x_1) = f'(c) (x_2 - x_1) $$

Since $f’(x) = 0$ for all $x$, then $f’(c) = 0$ and therefore

$$ f(x_2) - f(x_1) = 0 \leftrightarrow f(x_2) = f(x_1) $$

Thus $f$ has the same value at any two numbers $x_1$ and $x_2$ in $(a, b)$. This means that $f$ is constant on $(a, b)$.

PROOF: Let $F(x) = f(x) - g(x)$. Then

$$ F'(x) = f'(x) - g'(x) = 0 $$

for all $x \in (a, b)$. Thus by the previous theorem $F$ is constant, that is $f - g$ is constant.

This corollary says that if two functions have the same derivatives on an interval, then their graphs must be vertical trasnlation of each other.

What Derivatives Tell Us about the Shape of a Graph

What Does $f’$ say about $f$?

See Figure 1 for a graphical representation.

PROOF: Without loss of generality we let $x_1$ and $x_2$ be any two numbers on 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 $[x_1, x_2]$. So, by the Mean Value Theorem, there is a number $c$ between $x_1$ and $x_2$ such that

$$ f(x_2) - f(x_1) = f'(c)(x_2 - x_1) $$

We know that $f’(c) > 0$ by our initial assumption and $x_2 - x_1$ because $x_1 < x_2$, thus the right side of the previous equation must be positive:

$$ f(x_2) - f(x_1) > 0 $$

Which is equivalent to

$$ f(x_2) > f(x_1) $$

This shows that $f$ is increasing.

The First Derivative Test

The First Derivative Test is illustrated in Figure 4.

What does $f’’$ say about $f’$?

See Figure 7 for an illustration on the concavity of a function.

In Figure 7(a) we see that the slope of the tangent increases. This means that the derivative $f’$ is an increasing function and therefore its derivative $f’’$ is positive. Likewise, in Figure 7(b) the slope of the tangent decreases, so $f’$ decreases and therefore $f’’$ is negative.

The Second Derivative Test

Note that The Second Derivative Test is inconclusive when $f’’(c) = 0$ as there might be a maximum, a minimum or neither in that point. This test also fails when $f’’(c)$ does not exist.

Indeterminate Forms and l’Hospital’s Rule

Indeterminate Forms (Types $\frac{\infty}{\infty}$, $\frac{0}{0}$)

In general, if we have a limit of the form:

$$ \lim_{x \to a} \frac{f(x)}{g(x)} $$

where both $f(x) \to 0$ and $g(x) \to 0$ as $x \to a$, then this limit may or may not exist, and is called an indeterminate form of type $\frac{0}{0}$.

In general, if we have a limit of the form:

$$ \lim_{x \to a} \frac{f(a)}{g(a)} $$

where both $f(x) \to \infty$ (or $-\infty$) and $g(x) \to \infty$ (or $-\infty$) as $x \to a$, then this limit may or may not exist, and is called an indeterminate form of type $\frac{\infty}{\infty}$.

L’Hospital’s Rule

We introduce a systematic method for the evaluation of indeterminate forms of type $\frac{0}{0}$ or type $\frac{\infty}{\infty}$.

Note:

1. L’Hospital’s Rule says that the limit of a quotient of function is equal to the limit of the quotient of their derivatives, provided that the conditions are satisfied.
2. L’Hospital’s Rule is also valid for one-sided limit, and for limits at infinity or negative infinity.
3. For the special case in which $f(a) = g(a) = 0$, $f’$ and $g’$ are continuous and $g’(a) \neq 0$, it is easy to show why l’Hospital’s Rule is true:

Given $f’$ and $g’$ are continuous:

$$ \lim_{x \to a} \frac{f'(x)}{g'(x)} = \frac{f'(a)}{g'(a)} $$

By the definition of a derivative:

$$ \frac{f'(a)}{g'(a)} = \frac{\lim_{x \to a} \frac{f(x) - f(a)}{x - a}}{\lim_{x \to a} \frac{g(x) - g(a)}{x - a}} $$

By the properties of limits:

$$ \frac{\lim_{x \to a} \frac{f(x) - f(a)}{x - a}}{\lim_{x \to a} \frac{g(x) - g(a)}{x - a}} = \lim_{x \to a} \frac{\frac{f(x) - f(a)}{x - a}}{ \frac{g(x) - g(a)}{x - a}} $$$$ \lim_{x \to a} \frac{\frac{f(x) - f(a)}{x - a}}{ \frac{g(x) - g(a)}{x - a}} = \lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)} $$

Because $f(a) = g(a) = 0$

$$ \lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)} = \lim_{x \to a} \frac{f(x)}{g(x)} $$

It is more difficult to prove the general version of l’Hospital’s Rule.

Indeterminate Products (Type $0 \cdot \infty$)

If $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = \infty$ or ($- \infty$), then it isn’t clear what the value of $\lim_{x \to a} [f(x)g(x)]$ will be.

This kind of limit is called an indeterminate form of type $0 \cdot \infty$. We can deal with it by writing the product $fg$ as a quotient:

$$ fg = \frac{f}{\frac{1}{g}} \text{ or } fg = \frac{g}{\frac{1}{f}} $$

This converts the given limit into an indeterminate form of type $\frac{0}{0}$ or $\frac{\infty}{\infty}$ so that we can use l’Hospital’s Rule.

Indeterminate Differences (Type $\infty - \infty$)

If $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = \infty$, then the limit:

$$ \lim_{x \to a} [f(x) - g(x)] $$

is called an indeterminate form of type $\infty - \infty$. There is a contest between $f$ and $g$. Will the answer be $\infty$ ($f$ wins), $-\infty$ ($g$ wins) or will they compromise on a finite number? To find out, we try to convert the difference into a quotient so that we can have an indeterminate form of type $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

Indeterminate Powers (Types $0^0, \infty^0, 1^{\infty}$)

Several indeterminate forms arise from the limit:

$$ \lim_{x \to a} [f(x)]^{g(x)} $$

1. $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$, type $0^0$
2. $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = 0$, type $\infty^0$
3. $\lim_{x \to a} f(x) = 1$ and $\lim_{x \to a} g(x) = \pm \infty$, type $1^{\infty}$

Each of these three cases can be treated by taking the natural logarithm as follows:

$$ \text{let } y = [f(x)]^{g(x)} \text{, then } \ln y = g(x) \ln f(x) $$

Now we calculate:

$$ \lim_{x \to a} g(x) \ln f(x) $$

This limit is always an indeterminate product (indeterminate form of type $0 \cdot \infty$), and can be solved as one, which will require the application of L’Hospital’s Rule.

Once we get the value of the limit for $\ln y$, we can obtain the limit of $y$ by writing the logarithm of $y$ as an exponential:

$$ \lim_{x \to a} y = \lim_{x \to a} [f(x)]^{g(x)} = \lim_{x \to a} e^{\ln y} = \lim_{x \to a} e^{g(x) \ln f(x)} = e^{\lim_{x \to a} g(x) \ln f(x)} $$

Summary of Curve Sketching

Guidelines for Sketching a Curve

The following checklist is intended as a guide to sketching a curve $y = f(x)$.

1. Domain: Start by determining the domain $D$ of $f$.
2. Intercepts:
   • Find where the curve intersects the $y$-axis by evaluating $f(0)$.
   • Find where the curve intersects the $x$-axis by solving $f(x) = 0$ for $x$.
3. Symmetry:
   • If $f(x) = f(-x)$ for all $x \in D$, then $f$ is an even function and the curve is symmetric about the $y$-axis.
   • If $-f(x) = f(-x)$ for all $x \in D$, then $f$ is an odd function and the curve is symmetric about the origin.
   • If $f(x) = f(x + p)$ for all $x \in D$, where $p$ is a positive constant then $f$ is a periodic function and the smallest such number $p$ is called the period.
4. Asymptotes:
   • Horizontal asymptotes: If either $\lim_{x \to \infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$, then the line $y = L$ is a horizontal asymptote of the curve $y = f(x)$.
   • Vertical asymptotes: The line $x = a$ is a vertical asymptote if either $\lim_{x \to a^+} f(x) = \infty$ and $\lim_{x \to a^-} f(x) = \infty$ or $\lim_{x \to a^+} f(x) = -\infty$ and $\lim_{x \to a^-} f(x) = -\infty$.
   • Slant asymptotes: these will be discussed later on.
5. Intervals of Increase or Decrease: Use the Increasing/Decreasing Test to find the intervals in which $f$ increases or decreases.
6. Local Maximum or Minimum Values: Find the critical number of $f$, then use the First Derivative Test to determine whether the critical numbers are maximum or minimum values for $f$.
7. Concavity and Points of Inflection: Compute $f’’(x)$ and use the Concavity Test to determine where $f$ is concave upward and where $f$ is concave downward.
8. Sketch the Curve: Using the information from the previous points, draw the graph of $f$.

Slant Asymptotes

Some curve have asymptotes that are oblique. If:

$$ \lim_{x \to \infty} [f(x) - (mx + b)] = 0 $$

where $m \neq 0$, then the line $y = mx + b$ is called a slant asymptote because the vertical distance between the curve $y = f(x)$ and the line $y = mx + b$ approaches $0$ (see Figure 12). A similar situation exists is we let $x \to - \infty$.

In the case of rational functions, slant asymptotes occur when the degree of the numerator is one more than the degree of the denominator. In such a case the equation of the slant asymptote can be found by long division.

Example

Given a function $f(x) = xe^{-\frac{1}{x}}$, we obtain its slant asymptote, denoted by $y = mx + b$ by first finding the slope $m$ as:

$$ m = \lim_{x \to \pm \infty} \frac{f(x)}{x} = \frac{xe^{-\frac{1}{x}}}{x} = e^{-\frac{1}{x}} = e^{0} = 1 $$

Thus $m = 1$. Now we compute the value for the intercept $b$ as:

$$ b = \lim_{x \to \pm \infty} (f(x) - mx) = xe^{-\frac{1}{x}} - x = x \left(e^{-\frac{1}{x}} - 1\right) = \infty \cdot (1 - 1) = \infty \cdot 0 $$

We apply L’Hopital’s Rule:

$$ = \lim_{x \to \pm \infty} \frac{e^{- \frac{1}{x}} - 1}{\frac{1}{x}} = -\frac{e^{- \frac{1}{x}}x^{-2}}{x^{-2}} = -e^{- \frac{1}{x}} = e^{0} = -1 $$

Such that the slant asymptote is given by $y = mx + b = x - 1$.

Practical questions

How to prove that a function is not periodic

A function $f(x)$ is periodic if it exists $T > 0$ such that:

$$ f(x + T) = f(x) \forall x $$

There are a few ways in which we could prove $f$ is not periodic:

1. For simple functions you can directly check whether a period $T > 0$ can exist by solving:

$$ f(x + T) = f(x) $$

2. Find a contradiction with specific values.
   • Pick a generic $x_1$ and try to find a generic $T$ that satisfies $f(x + T) = f(x)$, if there is not such $T$, then $f$ is not periodic.
   • If there is such a $T$, repeat the previous step with another generic $x_2$, repeat until you can disprove the periodicity (try with a sensible number of points).
3. If the function increases or decreases without bound (i.e. it tends to $\pm \infty$), then we can assume there is no periodicity because the values of $f$ keep getting larger and larger and do not follow a pattern.
4. If it is a composite function, where one of the components is a non-periodic function, then the composite function is also non-periodic.

How to prove that a function does not have asymptotes

To show there exist no vertical asymptotes for a function $f(x)$ we need to check limits near critical points:

• For a rational function, find $x$ such that the denominator is $0$.
• For logarithms, the argument of the function must be postive. So boundary points like $x = 0, \text{for} \ln(x)$ are candidates.
• For square roots or even roots, its argument must be non-negative. So boundary points like $x < 0 \text{for} \sqrt{x}$ are candidates.
• Any other undefined points.

This is to say that if

1. If $f$ is defined everywhere, then there is no vertical asymptote.
2. If the limits on critical points where $f$ would be undefined are finite, than there is no vertical asymptote.

To show there exist no horizontal asymptotes for a function $f(x)$ we have to study how the function behaves when $x$ increases or decreases without bound. If $f$ is infinite or does not exist when $x$ increases or decreases without bound, then there are no horizontal asymptotes.

To show there exist no oblique asymptotes for a function $f(x)$ we must show that the limit:

$$ \lim_{x \to \infty} \frac{f(x)}{x} = m $$

or

$$ \lim_{x \to \infty} (f(x) - mx) = b $$

either do not exist, or one of them is infinite.

Optimization Problems

Proof (Absolute Minimum)

Suppose $f$ is a continuous differentiable function on an interval $I$, and let $c \in I$ be a critical point of $f$. Let us assume:

$$ \forall x \in I, \text{ where } x < c, f'(x) < 0 \text{ and } \forall x \in I, \text{ where } x > c, f'(x) > 0 $$

(1) Given any $x \in I$ with $x < c$, by The Mean Value Theorem there exists some $\xi \in (x, c)$ such that

$$ f'(\xi) = \frac{f(c) - f(x)}{c - x} $$

Because $\xi < c$ we know that $f’(\xi) < 0$ by assumption. Also $c > x \leftrightarrow c - x > 0$. Hence

$$ f(c) - f(x) < 0 \leftrightarrow f(c) < f(x) $$

That is, for every point $x$ to the left of $c$, $f(x)$ is bigger than $f(c)$.

(2) Given any $x \in I$ with $x > c$, by The Mean Value Theorem there exists some $\nu \in (c, x)$ such that

$$ f'(\nu) = \frac{f(x) - f(c)}{x - c} $$

Because $\nu > c$ we know that $f’(\nu) > 0$ by assumption. Also $x > c \leftrightarrow x - c > 0$. Hence

$$ f(x) - f(c) > 0 \leftrightarrow f(x) > f(x) $$

That is, for every point $x$ to the right of $c$, $f(x)$ is bigger than $f(c)$.

(3) By combining (1) and (2) we show that for all points on the interval $I$ that are not $c$, its value under $f$ is bigger than $f(c)$ thus we conclude that $f(c)$ is the abolute minimum of $f$ on the interval $I$.