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# Preview of calculus: part 2

### From Mathematics Is A Science

This is a part of Calculus 1: course.

## Contents

- 1 Piecewise Defined Functions
- 2 Characteristics of Functions
- 3 Growth: Increasing/Decreasing Behavior
- 4 Word Problems
- 5 We need more functions
- 6 Linear Functions
- 7 Quadratic Functions
- 8 Power Functions:
- 9 Rational Functions
- 10 Algebraic Functions
- 11 Trigonometric Functions
- 12 New Functions From Old
- 13 Compositions:
- 14 Algebra of Functions

## 1 Piecewise Defined Functions

The absolute value function $$f(x) = |x|$$ is computed as follows: $$f(x) = \text{ if } x < 0 \text{ then } y = - x $$ $$f(x) = \text{ if } x \geq 0 \text{ then } y = x$$

We can re-write this algebraically as $$ f(x) = \begin{cases} -x & \text{ if } x < 0 \\ x & \text{ if } x \geq 0 \end{cases} $$

This is called a *piecewise* definition. It's a special case of algebraic representation of functions.

Where do functions like this comes from?

**Example:**

Hypothetically...

The law says \( \begin{cases} \text{if income } & < 10000 \text{ then tax rate } = 0\% \\ \text{if } 10000 & \leq \text{ income } < 20000 \text{ then tax rate } = 10\% \\ \text{if income } & \geq 20000 \text{ then tax rate } = 20\% \end{cases} \)

We can express this algebraically.

Suppose \( x \) is the income and \( y = f(x) \) is the tax rate, then $$ f(x) = \begin{cases} 0 & \text{if } x < 10000 \\ 10 & \text{if } 10000 < x < 20000 \\ 20 & \text{if } x \geq 20000 \end{cases} $$

There is a possible problem with the above. There is no definition for \( x = 10000 \)! Fix it.

## 2 Characteristics of Functions

Start with symmetry.

Functions: | Even | Odd | Neither |
---|---|---|---|

Graphical | |||

Algebraic | \( y = x^{2} \) | \( y = x^{3} \) | $y=x+1$, just about anything... |

| |||

We discover | Same \( y \) for two different \( x \)'s (\( x \) & \(-x\)). | \( y \) for \( x \) and \( -y \) for \(-x\). | |

Definition: |
\( f(x) = f(-x) \) | \( f(-x) = -f(x) \) | |

Examples | \( (-x)^{2} = x^{2} => y = x^{2} \) is even | \( (-x)^{3} = -x^{3} => y = x^{3} \) is odd | |

Other Examples | \( \cos x, x^{4}, x^{6}, x^{6} + x^{4} - 17x^{2} \). | \( x, x^{3}, x^{5}, x^{7}, x + x^{5} - 17 x^{17}, \sin x \). | |

Numerical | Similar | ||

Warning: there could be more values and these observations could turn out to be false:

## 3 Growth: Increasing/Decreasing Behavior

This verbal definition is simple and the geometric meaning is very clear. However, both are imprecise. Let's work out this concept algebraically.

Function \( y=f(x) \) is *increasing on interval* \( (a, b) \) if for any given \( x_{1}, x_{2} \) in \( (a, b) \),

$$ \begin{array}% \text{ if } x_{1} < x_{2} & \text{ then } & f(x_{1}) < f(x_{2}), \\ \end{array} $$

How do we verify these conditions? It's hard (but easy with calculus).

**Example:**

$$ f(x) = 3x - 7 $$

If \( x_{1} < x_{2} \) then

$$ \begin{aligned} f(x_{1}) = 3 x_{1} - 7 & \overset{?}{<} f(x_{2}) = 3 x_{2} - 7 \\ 3 x_{1} & \overset{?}{<} 3 x_{2} \\ \therefore x_{1} & < x_{2} \end{aligned} $$

The computation suggests that $y=f(x)$ is increasing. To finish the proof, retrace your steps.

This is even harder for quadratic, cubic, ... functions as they lead to quadratic, cubic, .... equations.

Function \( y=f(x) \) is *decreasing on interval* \( (a, b) \) if for any given \( x_{1}, x_{2} \) in \( (a, b) \),

$$ \begin{array}% \text{ if } x_{1} < x_{2} & \text{ then } & f(x_{1}) > f(x_{2}), \\ \end{array} $$

Notation:

- $\nearrow $ for increasing, and
- $\searrow $ for decreasing.

**Review exercise:**

$$ y = \sqrt{u} + \sqrt{4 - u} $$

Find the domain, i.e. find all \( u \)'s for which the formula makes sense.

- Start
- Make sure that \( \sqrt{} \) makes sense, i.e. what's inside can not be negative.

- \( \sqrt{u} \Rightarrow u \geq 0 \), solved already
- \( \sqrt{4 - u} \Rightarrow 4 - u \), solve

- Subtract 4 from both sides, \( -u \geq -4 \)
- Multiply by \( (-1) \), \( u \leq 4 \)

Each \( u \) in the domain must satisfy both inequalities, \( u \geq 0 \) **AND** \( u \leq 4 \). Hence

$$ A: D = [0,4] $$

## 4 Word Problems

**Problem.** A farmer with 100 yards of fencing material wants to build as large a rectangular enclosure as possible.

He would first try to apply his common sense.

**Trial and Error:**

$$ \begin{aligned} 25 \times 25 & = 625 \text{ sq. yards} \\ 24 \times 26 & = 624 \ldots \\ \vdots & = \vdots \vdots \end{aligned} $$

More trials lead to better results but we cannot try all possible rectangles.

**Numerical analysis with Excel:**

Start with the perimeter : \( 2( W + D ) = 100\), where $W,D$ are the width and the depth of the rectangle.

Solve it $$ D = 50 - W $$ What about the area? $$ \begin{aligned} A & = WD \\ & = W(50-W) \end{aligned} $$

Using Excel, we see this:

Excel fills gaps with straight lines. We can see that the graphical representation is better and it suggests that 25 is the best solution.

But gaps are still an issue.

**Algebra:**

We can restate the problem as $$ \begin{aligned} A & = W( 50 - W ) \\ & \textrm{Find the maximum} \end{aligned} $$

Later, we will use the derivative. Now we use the fact that this is a parabola.

What do we know about it?

- There is one max or min.
- But which one?

\( f(x) = -x^{2} + 50 x \) (We replaced \( W \) with \( x \) here.)

Because of the "-", this one opens *down*, with a max.

**Question.** Where is the maximum?

**Answer.** It is the midpoint between the two \( x \)-intercepts.

In our case, the \( x \)-intercepts are 0 and 50. So the "vertex" of the parabola is at

$$ x = \frac{ 0 + 50 }{ 2 } = 25 $$

We've solved the problem but it's clear that with even a bit more complicated functions we wouldn't be able to handle the algebra. Calculus will help...

## 5 We need more functions

We need a library of elementary functions to build others.

A parabola (curve) is the graph of a function. But what kind of function?

**Quadratic Functions**

$$ \begin{aligned} & x^{2} \\ & x^{2} + 1 \\ & x^{2} + x + 1 \end{aligned} $$

**Example:**

Is this a parabola?

We don't know. Probably not because it has a flat bottom.

In fact, it looks like \( y = x^{4} \), which is not a parabola.

## 6 Linear Functions

Linear functions are simpler than quadratic functions.

Slope-intercept form: \( \begin{matrix} y = & mx & b \\ & \uparrow & \uparrow \\ & \textrm{the slope} & y\textrm{-intercept} \end{matrix} \)

** Monotinicity is very simple:**

- \( m > 0 \) - \( f \) is increasing, on the whole domain.
- \( m < 0 \) - \( f \) is decreasing, on the whole domain.
- \( m = 0 \) - \( f \) is constant, on the whole domain (technically not "linear").

Also, there are no max/min points.

## 7 Quadratic Functions

The general form is $$ f(x) = ax^{2} + bx + c $$

Facts we know

- \( a > 0 \)
- parabola opens up

- \( a < 0 \)
- parabola opens down

- \( a = 0\)
- It's not "quadratic".

The x-coordinate of the vertex of parabola (i.e., max or min) is $$ v = - \frac{b}{2 a } $$ Find y.

Domain is all reals.

Now, a quadratic function will exhibit either:

- \( \underbrace{(-\infty, v)}_{\nearrow} \) and \( \underbrace{(v, \infty)}_{\searrow} \) - Maximum
- \( \underbrace{(-\infty, v)}_{\searrow} \) and \( \underbrace{(v, \infty)}_{\nearrow} \) - Minimum

Polynomials are "made up" of powers of \( x \) as follows.

## 8 Power Functions:

$$ x^{0} = 1 , \underbrace{x}_{\text{linear}}, \underbrace{x^{2}}_{\text{quadratic}}, \underbrace{x^{3}}_{\text{cubic}}, \cdots , x^{n}, \cdots $$

The degree, odd vs even, affects the shape of the graph.

Power is even, then it looks "like" \( x^{2} \) .

Power is odd, then it and looks "like" \( x^{3} \) .

We can think algebraically and justify the pictures.

Later, we'll use the derivative.

\( f(x) = 5x^{3} - 3x^{2} + 17x - 18 \), cubic

**Fact:** The domain is all real numbers (no division, etc).

## 9 Rational Functions

Called this way because they are *ratios* of polynomials.

$$ \left. \frac{ x - 1 }{ x + 1 } \right\} \text{ polynomials } $$

Also

$$ \left(\frac{1 + x}{x^{2}}\right) \frac{x^{3}}{x^{2} + 1} $$

**Definition:** Function $f$ is called *rational* if

$$ f(x) = \frac{P(x)}{Q(x)} $$

where \( P \) and \( Q \) are polynomials.

**Question:**

What is the domain of a rational function?

**Example:**

What is the domain of \( \dfrac{x - 1}{x + 1} \)?

Answer: $x \neq -1 $

I.e., the $f(x)$ makes sense for all \( x \) but -1.

**Solution:**

We look at the denominator, \( x + 1 = 0 \) and solve.

$$ \Longrightarrow x = -1 $$

not in the domain.

If we plug \( x = -1 \), we have a division by 0.

**Problem:**

What is the domain for

$$ f(x) = \frac{x - 1}{x^{2} - 4} $$

Set the denominator to 0 and solve

$$ \begin{aligned} x^{2} - 4 & = 0 \\ x^{2} & = 4 \\ x & = \pm 2 \end{aligned} $$

So the domain consists of all real numbers except \( \pm 2 \).

## 10 Algebraic Functions

I.e., the ones "with roots".

**Problem.** Find the domain to the function

$$ \sqrt{ x - \sqrt{x}} $$

Whatever is inside the square root can't be negative. So,

- \( x \geq 0 \), i.e. \( x \) is positive.
- \( x - \sqrt{x} \geq 0 \), solve it. $$ \begin{aligned}x & \geq \sqrt{x} \\ x^{2} & \geq x \\x & \geq 1\end{aligned}$$

The domain is

$$D = [1, \infty). $$

## 11 Trigonometric Functions

These three will suffice 00% of the time: $$ \sin x, \cos x, \tan x = \frac{\sin x}{\cos x} $$

**Fact:** The domain for sin and cos are all real numbers.

**Fact:** These functions are periodic, with period $2\pi$.

$$ \begin{aligned} \sin( x + 2 \pi ) & = \sin x \\ \cos(x + 2\pi) & = \cos x \\ \tan(x+2\pi) &= \tan x \end{aligned} $$

for all \(x\).

Recall the definitions (comes from the Pythagorean Theorem):

**Domain of \(\tan x \)?**

$$ \tan x = \frac{ \sin x}{\cos x}$$

To find domain, set denominator to 0 and solve. Find all \( x \)'s

$$ \begin{aligned} \cos x & = 0 \\ x & = \ldots, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots \end{aligned} $$

**Question:**

Is it increasing? It looks like it's increasing everywhere... But the correct question is: on what intervals is it increasing? In particulr, is it increasing on \( (-\infty, \infty) \)?

No. Because \( \tan \) is undefined at \( \dfrac{\pi}{2}, \dfrac{3\pi}{2}, \ldots \) in the first place. Plus, you can see the jumps.

**Review exercise:**

Plot the graph of the piece-wise defined function:

$$ f(x) = \begin{cases} 3 - \frac{1}{2}x & \text{ for } x \leq 2 \\ 2x - 5 & \text{ for } x > 2 \end{cases} $$

**Plot:**

- Plot \( 3 - \dfrac{1}{2} x \).
- Plot \( 3 - \dfrac{1}{2}x \) with domain \( x \leq 2 \).
- Plot \( 2x - 5 \).
- Plot \( 2x - 5 \) with domain \( x > 2 \).
- Combine 2) and 4).

**Answer:**

## 12 New Functions From Old

We set up a library of elementary functions: powers, roots, exponent, sin, cos. Then we combine them algebraically with the 4 operations (+ one more). For example:

$$ x^{3}, x^{5} \rightarrow x^{3} + x^{5} $$

The idea is below:

Graphical representations: How do these transformations of graphs affect the functions?

- Shift (Two types: Horizontal and Vertical)
- Stretch (Two types: Horizontal and Vertical)
- Flip (Two types: Horizontal and Vertical)

Six in Total

**Shift (Vertical)**

**Rule 1:**

If the graph \( y = g(x) \) is that of \( y = f(x) \) shifted \( k \) units up, then $$ g(x) = f(x) + k $$ This is what happens to every point on the graph:

Fact: Same exact shape of the graph.

**Example:**
$$\begin{aligned}x^{2}, x^{2} + 1, x^{2} + 10, & \textrm{ same shape} \\x^{2} - 1 , x^{2} - 10, & \textrm{ same shape}\end{aligned}$$
What if we shift down? We have to rewrite the rule?

**Rule 1 (rewritten):**

If the graph \( y = g(x) \) is that of \( y = f(x) \) shifted \( k \) units up/down, then $$ g(x) = f(x) \pm k $$ No need for "down". Indeed, "-10 units up" mean "10 units down".

**Example:**

Suppose "I drive West, drive one hour at -60 m/h." Where am I? A: 60 miles east of here. $$ \begin{aligned}\textrm{distance} & = \textrm{time} \times \textrm{velocity} \\& = (1) \times (-60) \\& = -60\end{aligned}$$ What about the horizontal shift? Compare.

**Vertical Shift:** \( y = f(x) \Rightarrow y = f(x) + k \)

$$ \therefore y_{1} = y_{0} + k,$$ So $$ g(x) = f(x) + k $$

**Horizontal Shift:**

$$ y_{0} = f(x_{0}) = g(x_{0} + k), $$ So $$ g(x) = f(x - k) $$

**Example:**

\( f(x) = x^{2} \), shift 2 units to the right and we get $$ g(x) = f(x-2) = (x-2)^{2} $$ \( x \) changes.

What is the \( x \) -intercept of \( g \)?

Set \( f(x) = 0 \), and solve. \( \begin{array}{c | c}( x + 2) ^{2} = 0 & \textrm{opposite?} \\x - 2 = 0 & h(x) = (x + 2)^{2} \\x = 2 & x + 2 = 0 \\& x = -2\end{array}\) \( h(x) \) is shifted 2 units to the left.

**Rule 1-2:**

If the graph of \( g \) is that of \( f \), shifted \( k \) units to the right, then
$$ g(x) = f(x-k)$$
**Algorithmic interpretation of these shifts.**

**Vertical Shift**

**Horizontal Shift**

**Example:**

Let \( h(x) = (x + 2)^{2} - 3 \)
**What is the graph?**

A different kind of transformation.

**Flip**

What about the algebraic representation?

Then \( y_{1} = - y_{0} \), so $$ g(x) = - f(x) $$

**Rule 2-1:**

If the graph of \( g \) is that of \(f \), flipped about the \( x \)-axis then $$ g(x) = - f(x) $$

**Example:**

$$ h(x) = - f(x) + 3 $$ How do we get the graph of \( h \)? $$ x \longmapsto f(x) \underbrace{\longmapsto}_{\textrm{vertical flip}} - f(x) \underbrace{\longmapsto}_{\textrm{vertical shift}} - f(x) + 3 $$ The order matters... To know the order, read the function from inside-out.

**Rule 2-2**

If the graph of \( y = g(x) \) is that of \( y = f(x) \) flipped about the \( y \)- axis, then $$ g(x) = f(-x) $$

**Stretch:**

The graph is not a wireframe anymore! The \( xy \)-plane is the one stretched, if you think of it as a rubber sheet, with a graph on it.

**Consider the stretch by a factor of 2**

**Rule 3-1**

If the graph of \( y = g(x) \) is that of \( y = f(x) \) stretched vertically by a factor of \( k > 0 \), then \( g(x) = kf(x) \). (Factor outside of \( k \) ).

**Review exercise:** Match graphs and formulas.

**What about horizontal stretch?**

**Rule 3-2**

If graph \( y = g(x) \) is that of \( y = f(x) \) stretched by the factor \( k \)-horizontally, then \( g(x) = f(x/k) \). (Factor inside of \( f \) ).

**Example:**
Given
$$ f(x) = x $$
Stretched by 2x vertically is
$$ g(x) = 2 x $$
On the other hand:

We can get \( y = g(x) \) by shrinking \( y = f(x) \) horizontally, factor 2x. $$ 2x = \frac{x}{\frac{1}{2}} $$ Stretch by factor \( \frac{1}{2}\) = shrink by 2.

**Example:**

$$ f(x) = x^{2} $$ Vertical stretch by 2 is \( g(x) = 2x^{2} \).

Horizontal stretch by \( \sqrt{2} \) is \( g(x) = (\sqrt{2} x )^{2} \). $$ \begin{aligned}(\sqrt{2} x)^{2} & = (\sqrt{2})^{2} \cdot x^{2} \\ & 2 x^{2}\end{aligned}$$

**Example:**

Verify the plot \( 2 x^{2} \) vs. \( (2 x )^{2} \).

Not the same!

More generally we interpret these operations as compositions.

## 13 Compositions:

Consider:

**More Generally:**

Variables have to match!

**Example:**

Represent \( z = h(x) = \sqrt[3]{x^{2} + 1 } \) as the composition of two functions. (Easy if you know how). $$ x \rightarrow x^{2} + 1 \xrightarrow{y} \sqrt[3]{y} \xrightarrow{z} $$

**Example:**

Given Speed (60 m/h), Miles per gallon (30 m/gal) and Cost per gallon ($5/gal), represent the expense as a function of time.

*Analysis:*

$$ \text{time (hours)} \xrightarrow{\text{60 m/h}} \text{distance (miles)}\xrightarrow{\text{30 m/gal}} \text{gas used (gal)} \xrightarrow{\$\text{5/gal}}\text{expense } (\$) $$

Composition:

$$ t \rightarrow \underset{f}{ 60 t } \rightarrow d \rightarrow \underset{k}{\frac{d}{30}} \rightarrow g \rightarrow \underset{h}{5 g} \rightarrow e $$

Compute:

$$ d = f(t) = 60t, \; g = k(d) = \frac{d}{30} \; e = h(g) = 5g $$

Substitute $$ \begin{aligned}g = k(d) & = k(f(t)) \\e = h(g) & = h(k(d)) \\& = h(k(f(t)))\end{aligned}$$ Read from right to left. ( \( f(t) \) is not \( (t)f \).

**Example:**
\( f(x) = x^{2} \) and \( g(x) = \cos x \).
Find \( f(g(x)) \), replace \( x \) in \(f \) with \( (\cos x ) \), always with parentheses.
$$ x^{2} \Longrightarrow ( \cos x )^{2} $$
or \( cos^{2} x \).

**Example:**

Find \( g(f(x)) \). Replace \( x \) in \( g \) with \( (x^{2}) \). $$ \cos x \rightarrow \cos (x^{2}) $$

**Review Exercise.**

- Stretch vertically by 2 (about \( y \): \( g(x) = 2 f(x) \) ).
- Shift horizontally - right by 2 (about \( x \): \( h(x) = g(x-2) \) ).Replace \( x \) with \( ( x-2 ) \).(Suggestion: use parentheses a lot).

*Algebraically:*

$$ f(x) = \sqrt{3 x - x^{2}} \Longrightarrow ?????\sqrt{3x - x^{2} - 2} $$

Instead, substitute:

$$ \sqrt{3 ( x - 2 ) - ( x - 2 )^{2} } $$

## 14 Algebra of Functions

For opertions, we need to align the variables first -- but in a different way.

**Example:**
Find the composition of these functions.
$$ f(x) = x + 1, \; g(x) = 2x, \; h(x) = x - 1. $$
Find \( f \circ g \circ h \).

Rename the variable to make them match:

$$ y = h(x) = x - 1, \; z = g(y) = 2y, \; u = f(z) = z + 1 $$

Algorithm:

$$ \begin{alignat}{3}& & y = x - 1 \\& z = 2y & \\u = z + 1 & &\end{alignat}$$

Algebra:

$$\begin{aligned}u = z + 1 & = (2y) + 1 \\& = 2y + 1 \\& = 2(x - 1) + 1\end{aligned}$$