Calculus (Math 1A) Lecture 2vivek/1A/2.handout.pdf · the function f given by the formula x 7!x2 +...

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Calculus (Math 1A) Lecture 2 Vivek Shende August 26, 2017

Transcript of Calculus (Math 1A) Lecture 2vivek/1A/2.handout.pdf · the function f given by the formula x 7!x2 +...

Calculus (Math 1A)Lecture 2

Vivek Shende

August 26, 2017

Hello and welcome to class!

Last time

We saw various examples of functions and graphs; discusseddomain and range; increasing and decreasing functions, and evenand odd functions.

Today

We’ll review algebraic and trigonometric functions.

Constants

0

1

π +5

2

Constant functions

f (x) = 0

f (x) = 1

f (x) = π +5

2

Constant functions

It is not grammatical to say the function 1.

Instead you should say the function f (x) = 1 or the constantfunction with value 1.

Polynomials

0

1

x + 1

x2 + 3x + 5

x17 + 34x4 + 5x3 + 1

Polynomial functions

f (x) = 0

f (x) = 1

f (x) = x + 1

f (x) = x2 + 3x + 5

f (x) = x17 + 34x4 + 5x3 + 1

Polynomial functions

Again, it is not quite right to say the function x2 + 3x + 5.

Instead: the function f (x) = x2 + 3x + 5.

This itself is an abbreviation for something like:

the function f given by the formula x 7→ x2 + 3x + 5.

Linear functions

f (x) = ax + b

The graph of this is a line with slope a. I.e., take any two pointson the graph (x0, f (x0)) and (x1, f (x1)), then

f (x1)− f (x0)

x1 − x0=

(ax1 + b)− (ax0 + b)

x1 − x0=

ax1 − ax0x1 − x0

= a

One says this line has y-intercept b because it meets the y -axis atthe point (0, f (0) = b).

Linear functions

f (x) = −2x + 4

Linear functions

One often wants to find the (unique) linear function whose graphpasses through two given points (x0, y0) and (x1, y1). That is, weare looking for a linear f such that f (x0) = y0 and f (x1) = y1.

For linear f , the ratio f (x1)−f (x0)x1−x0

is the same for any x1, x0.

In particular, we can let one of the xi be a variable:

f (x)− f (x0)

x − x0=

f (x1)− f (x0)

x1 − x0=

y1 − y0x1 − x0

f (x) = (x − x0)

(y1 − y0x1 − x0

)+ f (x0)

Linear functions

Try it yourself: what linear f satisfies f (1) = 2 and f (3) = 4?

f (x)− f (1)

x − 1=

f (3)− f (1)

3− 1

and plugging in values

f (x)− 2

x − 1=

4− 2

3− 1= 1

f (x) = x + 1

Linear functions

What’s the range of the linear function f (x) = ax + b?

If a = 0, the range is just {b}.

Otherwise, the range is (−∞,∞). To see why this is, consider anynumber y ∈ (−∞,∞). Then y = ax + b can be solved for x ,giving x = (y − b)/a.

Since

y = a

(y − b

a

)+ b = f

(y − b

a

)we see that y is in the range.

Linear functions

It is also interesting to ask: what is the closest line to a collectionof n > 2 points?

We will learn later how to find such a line. It can always be done,but is not always meaningful.

Quadratic functions

f (x) = x2 − 5x + 10

What can you say about its graph?

Since f (0) = 10, the graph meets the y axis at (0, 10).

The roots are given by the quadratic formula as:

10±√

25− 40

2=

10±√−15

2

so there are no real roots, i.e., the graph does not meet the x axis.

Quadratic functions

f (x) = x2 − 5x + 10

For x � 0 or x � 0, one has f (x)� 0, since in these cases theterm x2 is positive and larger than the other terms.

Completing the square: writing

f (x) = x2 − 5x +25

4− 25

4+ 10 =

(x − 5

2

)2

+15

4

From this one can see f (x) ≥ 15/4, and the function achieves thisminimum value at x = 5/2.

Quadratic functions

f (x) = x2 − 5x + 10

Quadratic functions

f (x) = x2 − 5x + 10

The range looks to be [15/4,∞). Indeed, by completing thesquare, we showed that at least the range is contained in thisinterval. To actually show it is equal to this interval, we mustconvince ourselves that every value [15/4,∞) is actually achieved.

We could do this using the quadratic formula:f (x) = x2 − 5x + 10 = y has a solution exactly when25− 4(10− y) ≥ 0, i.e. when y ≥ 15/4.

Quadratic functions

f (x) = x2 − 5x + 10

Here’s a different sort of argument why the range is [15/4,∞). Weknow that 15/4 is in the range, and by plugging in large values ofx , we learn that arbitrarily large numbers are in the range as well.

Looking again at the graph, the function is not skipping values —so it must take on every value above 15/4.

This argument is not yet precise – what does it mean, not skippingvalues, and how do we actually know it is true of f (x)? We willmake it precise later using the notion of continuity.

Cubic functions

f (x) = x3

Cubic functions

f (x) = x3 + 2

Cubic functions

f (x) = x3 + x2 − x − 1

Cubic functions

f (x) = −x3 − x2 + 5x + 1

Cubic functions

What’s the range of a cubic?

The answer is: (−∞,∞). Intuitively, this is because: by choosinglarge positive and negative values of x , one can get arbitrarily largepositive and values of x3, which dominate the remaining terms ofthe cubic. By continuity, this means all values must be achieved.

In a week or two, we will make this argument precise.

Polynomial functions

f (x) = x5 − 3x4 − 5x3 + 15x2 + 4x − 12

Rational functions (polynomials divided by polynomials)

f (x) =x2 + 2x + 3

x2 − 1

Note this function is not defined at x ± 1, i.e. its range is(−∞,−1) ∪ (−1, 1) ∪ (1,∞).

Rational functions (polynomials divided by polynomials)

f (x) =x2 + 2x + 3

x2 − 1

Algebraic functions (now also roots)

f (x) =√x

Algebraic functions (now also roots)

f (x) =√x g(x) = 3√x h(x) = 4√x

Algebraic functions (now also roots)

f (x) =

√10− 1√

x − 2

Algebraic functions (now also roots)

f (x) =

√10− 1√

x − 2

What’s the domain?

First of all, we need x ≥ 2 for the inner square-root to be defined.But then in addition, we need 1/

√x − 2 ≤ 10, so actually,

x ≥ 2 + 1100 . The domain is [2.01,∞).

Algebraic functions (now also roots)

f (x) =

√10− 1√

x − 2

What’s the range?

The function only takes positive values. Plugging in 2.01 giveszero, and plugging in large values of x gives numbers increasinglyclose to

√10. . Looking at the graph, the function is increasing –

we will learn later how to argue more carefully that the function isincreasing! – and so the range is [0,

√10).

Algebraic functions (now also roots)

It’s not so easy to see the actual range from the previous graph:

Algebraic functions (now also roots)

After zooming in:

Trigonometric functions

Sine and cosine are defined by the following picture

We take tan(θ) = sin(θ)/cos(θ).

Trigonometric functions

Here are the graphs of these functions

Trigonometric functions

Trigonometric functions get used in describing phenomena whichare in any sense cyclical or periodic.

This includes both mechanical and natural constructions involvingactual circles and things like waves whose mathematicaldescriptions involve imaginary circles.