Midterm I Review

Review for Midterm I

Math 1a

October 21, 2007

Announcements

I Midterm I 10/24, Hall 7-9pm, Hall A and D

I Old exams and solutions on website

I problem sessions every night, extra MQC hours

Outline

LimitsConceptComputationLimits involving infinity

ContinuityConceptExamples

The Intermediate ValueTheorem

DerivativesConceptIntepretationsImplicationsComputation

Outline





The concept of LimitLearning Objectives

I state the informal definition of a limit (two- and one-sided)

I observe limits on a graph

I guess limits by algebraic manipulation

I guess limits by numerical information

Heuristic Definition of a Limit

DefinitionWe write

limx→a

f (x) = L

and say

“the limit of f (x), as x approaches a, equals L”

if we can make the values of f (x) arbitrarily close to L (as close toL as we like) by taking x to be sufficiently close to a (on either sideof a) but not equal to a.

Math 1a Midterm I Review - October 21, 2007.GWBSunday, Oct 21, 2007

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The error-tolerance game

This tolerance is too bigStill too bigThis looks goodSo does this

a

L



a

L



a

L


This tolerance is too big

Still too bigThis looks goodSo does this

a

L



a

L


This tolerance is too big

Still too big

This looks goodSo does this

a

L



a

L


This tolerance is too bigStill too big

This looks good

So does this

a

L


This tolerance is too bigStill too bigThis looks good

So does this

a

L

Outline





Computation of LimitsLearning Objectives

I know basic limits like limx→a x = a and limx→a c = c

I use the limit laws to compute elementary limits

I use algebra to simplify limits

I use the Squeeze Theorem to show a limit

Limit Laws

Suppose that c is a constant and the limits

limx→a

f (x) and limx→a

g(x)

exist. Then

1. limx→a

[f (x) + g(x)] = limx→a

f (x) + limx→a

g(x)

2. limx→a

[f (x)− g(x)] = limx→a

f (x)− limx→a

g(x)

3. limx→a

[cf (x)] = c limx→a

f (x)

4. limx→a

[f (x)g(x)] = limx→a

f (x) · limx→a

g(x)

Limit Laws, continued

5. limx→a

f (x)

g(x)=

limx→a

f (x)

limx→a

g(x), if lim

x→ag(x) 6= 0.

6. limx→a

[f (x)]n =[

limx→a

f (x)]n

(follows from 3 repeatedly)

7. limx→a

c = c

8. limx→a

x = a

9. limx→a

xn = an (follows from 6 and 8)

10. limx→a

n√

x = n√

a

11. limx→a

n√

f (x) = n

√limx→a

f (x) (If n is even, we must additionally

assume that limx→a

f (x) > 0)

Direct Substitution Property

Theorem (The Direct Substitution Property)

If f is a polynomial or a rational function and a is in the domain off , then

limx→a

f (x) = f (a)


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Theorem (The Squeeze/Sandwich/Pinching Theorem)

If f (x) ≤ g(x) ≤ h(x) when x is near a (as usual, except possiblyat a), and

limx→a

f (x) = limx→a

h(x) = L,

thenlimx→a

g(x) = L.


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Outline





Limits involving infinityLearning Objectives

I know vertical asymptotes and limits at the discontinuities of”famous” functions

I intuit limits at infinity by eyeballing the expression

I show limits at infinity by algebraic manipulation

DefinitionLet f be a function defined on some interval (a,∞). Then

limx→∞

f (x) = L

means that the values of f (x) can be made as close to L as welike, by taking x sufficiently large.

DefinitionThe line y = L is a called a horizontal asymptote of the curvey = f (x) if either

limx→∞

f (x) = L or limx→−∞

f (x) = L.

y = L is a horizontal line!

TheoremLet n be a positive integer. Then

I limx→∞1xn = 0

I limx→−∞1xn = 0

Using the limit laws to compute limits at ∞

Example

Find

limx→∞

2x3 + 3x + 1

4x3 + 5x2 + 7

if it exists.

A does not exist

B 1/2

C 0

D ∞

Using the limit laws to compute limits at ∞

Example

Find

limx→∞

2x3 + 3x + 1

4x3 + 5x2 + 7

if it exists.

A does not exist

B 1/2

C 0

D ∞

SolutionFactor out the largest power of x from the numerator anddenominator. We have

2x3 + 3x + 1

4x3 + 5x2 + 7=

x3(2 + 3/x2 + 1/x3)

x3(4 + 5/x + 7/x3)

limx→∞

2x3 + 3x + 1

4x3 + 5x2 + 7= lim

x→∞

2 + 3/x2 + 1/x3

4 + 5/x + 7/x3

=2 + 0 + 0

4 + 0 + 0=

1

2

Upshot

When finding limits of algebraic expressions at infinitely, look atthe highest degree terms.

SolutionFactor out the largest power of x from the numerator anddenominator. We have

2x3 + 3x + 1

4x3 + 5x2 + 7=

x3(2 + 3/x2 + 1/x3)

x3(4 + 5/x + 7/x3)

limx→∞

2x3 + 3x + 1

4x3 + 5x2 + 7= lim

x→∞

2 + 3/x2 + 1/x3

4 + 5/x + 7/x3

=2 + 0 + 0

4 + 0 + 0=

1

2

Upshot

When finding limits of algebraic expressions at infinitely, look atthe highest degree terms.

Infinite Limits

DefinitionThe notation

limx→a

f (x) =∞

means that the values of f (x) can be made arbitrarily large (aslarge as we please) by taking x sufficiently close to a but not equalto a.

DefinitionThe notation

limx→a

f (x) = −∞

means that the values of f (x) can be made arbitrarily largenegative (as large as we please) by taking x sufficiently close to abut not equal to a.

Of course we have definitions for left- and right-hand infinite limits.

Vertical Asymptotes

DefinitionThe line x = a is called a vertical asymptote of the curvey = f (x) if at least one of the following is true:

I limx→a f (x) =∞I limx→a+ f (x) =∞I limx→a− f (x) =∞

I limx→a f (x) = −∞I limx→a+ f (x) = −∞I limx→a− f (x) = −∞

Finding limits at trouble spots

Example

Let

f (t) =t2 + 2

t2 − 3t + 2

Find limt→a− f (t) and limt→a+ f (t) for each a at which f is notcontinuous.

SolutionThe denominator factors as (t − 1)(t − 2). We can record thesigns of the factors on the number line.

Finding limits at trouble spots

Example

Let

f (t) =t2 + 2

t2 − 3t + 2

Find limt→a− f (t) and limt→a+ f (t) for each a at which f is notcontinuous.

SolutionThe denominator factors as (t − 1)(t − 2). We can record thesigns of the factors on the number line.

(t − 1)−

1

0 +

(t − 2)−

2

0 +

(t2 + 2)+

f (t)1 2

+ ±∞ − ∓∞ +

(t − 1)−

1

0 +

(t − 2)−

2

0 +

(t2 + 2)+

f (t)1 2

+ ±∞ − ∓∞ +

(t − 1)−

1

0 +

(t − 2)−

2

0 +

(t2 + 2)+

f (t)1 2

+ ±∞ − ∓∞ +

(t − 1)−

1

0 +

(t − 2)−

2

0 +

(t2 + 2)+

f (t)1 2

+ ±∞ − ∓∞ +

(t − 1)−

1

0 +

(t − 2)−

2

0 +

(t2 + 2)+

f (t)1 2

+

±∞ − ∓∞ +

(t − 1)−

1

0 +

(t − 2)−

2

0 +

(t2 + 2)+

f (t)1 2

+ ±∞

− ∓∞ +

(t − 1)−

1

0 +

(t − 2)−

2

0 +

(t2 + 2)+

f (t)1 2

+ ±∞ −

∓∞ +

(t − 1)−

1

0 +

(t − 2)−

2

0 +

(t2 + 2)+

f (t)1 2

+ ±∞ − ∓∞

+

(t − 1)−

1

0 +

(t − 2)−

2

0 +

(t2 + 2)+

f (t)1 2

+ ±∞ − ∓∞ +


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Outline





Outline





ContinuityLearning Objectives

I intuitive notion of continuity

I definition of continuity at a point and on an interval

I ways a function can fail to be continuous at a point

Definition of Continuity

DefinitionLet f be a function defined near a. We say that f is continuous ata if

limx→a

f (x) = f (a).


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Free Theorems

Theorem

(a) Any polynomial is continuous everywhere; that is, it iscontinuous on R = (−∞,∞).

(b) Any rational function is continuous wherever it is defined; thatis, it is continuous on its domain.

Outline





The Limit Laws give Continuity Laws

TheoremIf f and g are continuous at a and c is a constant, then thefollowing functions are also continuous at a:

1. f + g

2. f − g

3. cf

4. fg

5. fg (if g(a) 6= 0)

Transcendental functions are continuous, too

TheoremThe following functions are continuous wherever they are defined:

1. sin, cos, tan, cot sec, csc

2. x 7→ ax , loga, ln

3. sin−1, tan−1, sec−1


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Math 1a Midterm I Review - October 21, 2007.GWBTuesday, Oct 23, 2007

Page16of25

Outline





The Intermediate Value TheoremLearning Objectives

I state IVT

I use IVT to show that a function takes a certain value

I use IVT to show that a certain equation has a solution

I reason with IVT

A Big Time Theorem

Theorem (The Intermediate Value Theorem)

Suppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.

Illustrating the IVT

Suppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.

x

f (x)

a b

f (a)

f (b)

N

cc1 c2 c3

Illustrating the IVTSuppose that f is continuous on the closed interval [a, b]

and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.

x

f (x)

a b

f (a)

f (b)

N

cc1 c2 c3

Illustrating the IVTSuppose that f is continuous on the closed interval [a, b]

and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.

x

f (x)

a b

f (a)

f (b)

N

cc1 c2 c3

Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) 6= f (b).

Thenthere exists a number c in (a, b) such that f (c) = N.

x

f (x)

a b

f (a)

f (b)

N

cc1 c2 c3

Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.

x

f (x)

a b

f (a)

f (b)

N

c

c1 c2 c3


x

f (x)

a b

f (a)

f (b)

N

cc1 c2 c3


x

f (x)

a b

f (a)

f (b)

N

c

c1 c2 c3

Using the IVT

Example

Prove that the square root of two exists.

Proof.Let f (x) = x2, a continuous function on [1, 2]. Note f (1) = 1 andf (2) = 4. Since 2 is between 1 and 4, there exists a point c in(1, 2) such that

f (c) = c2 = 2.

True or FalseAt one point in your life your height in inches equaled your weightin pounds.


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Outline





Outline





ConceptLearning Objectives

I state the definition of the derivative

I Given the formula for a function, find its derivative at a point“from scratch,” i.e., using the definition

I Given numerical data for a function, estimate its derivative ata point.

I given the formula for a function and a point on the graph ofthe function, find the (slope of, equation for) the tangent line

The definition

DefinitionLet f be a function and a a point in the domain of f . If the limit

f ′(a) = limh→0

f (a + h)− f (a)

h

exists, the function is said to be differentiable at a and f ′(a) isthe derivative of f at a.

Outline





The Derivative as a functionLearning Objectives

I given a function, find the derivative of that function fromscratch and give the domain of f’

I given a function, find its second derivative

I given the graph of a function, sketch the graph of itsderivative

Derivatives

TheoremIf f is differentiable at a, then f is continuous at a.

How can a function fail to be continuous?


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The second derivative

If f is a function, so is f ′, and we can seek its derivative.

f ′′ = (f ′)′

It measures the rate of change of the rate of change!


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Outline





Implications of the derivativeLearning objectives

I Given the graph of the derivative of a function...I determine where the function is increasing and decreasingI determine where the function is concave up and concave downI sketch the graph of the original function

I find and interpret inflection points


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Fact

I If f is increasing on (a, b), then f ′(x) ≥ 0 for all x in (a, b)

I If f is decreasing on (a, b), then f ′(x) ≤ 0 for all x in (a, b).

Fact

I If f ′(x) > 0 for all x in (a, b), then f is increasing on (a, b).

I If f ′(x) < 0 for all x in (a, b), then f is decreasing on (a, b).

Definition

I A function is called concave up on an interval if f ′ isincreasing on that interval.

I A function is called concave down on an interval if f ′ isdecreasing on that interval.

Fact

I If f is concave up on (a, b), then f ′′(x) ≥ 0 for all x in (a, b)

I If f is concave down on (a, b), then f ′′(x) ≤ 0 for all x in(a, b).

Fact

I If f ′′(x) > 0 for all x in (a, b), then f is concave up on (a, b).

I If f ′′(x) < 0 for all x in (a, b), then f is concave down on(a, b).

Outline





Computing DerivativesLearning Objectives

I the power rule

I the constant multiple rule

I the sum rule

I the difference rule

I derivative of x 7→ ex is ex (by definition of e)

Theorem (The Power Rule)

Let r be a real number. Then

d

dxx r = rx r−1

Rules for Differentiation

TheoremLet f and g be differentiable functions at a, and c a constant.Then

I (f + g)′(a) = f ′(a) + g ′(a)

I (cf )′(a) = cf ′(a)

It follows that we can differentiate all polynomials.

Rules for Differentiation

TheoremLet f and g be differentiable functions at a, and c a constant.Then

I (f + g)′(a) = f ′(a) + g ′(a)

I (cf )′(a) = cf ′(a)

It follows that we can differentiate all polynomials.

Derivatives of exponential functions

Factddx ex = ex

Midterm I Review

Technology

lim f x g x

continuedlim f x f x

limxa x

f x g x hx

limxa f x

values of f x

eitherlim f x

orlim f x