3B MAS

4. Functions

Limit of a FunctionGraphically the limiting value of a function f(x) as x gets closer and closer to a certain value (say 'a') is obtained by moving along the curve from both sides of 'a' as x moves toward 'a'.

The limiting value of f(x) as x gets closer and closer to 'a' is denoted by

x alim f (x)

Right/Left Hand LimitsAs x moves towards 'a' from right (left) hand side, the limiting value of f(x) is denoted by

)x(flimand)x(flimaxax

Limiting and Functional ValueIf both sides limits are equal,

Otherwise, does not exist.

Note that may not equal to f(a)

)x(flim)x(flim)x(flimaxaxax

)x(flimax

)x(flimax

a

f(x)

x

y

f(a)

)()(lim afxfax

Limiting Value

Example 1Find the limit of the f(x) as x approaches a for the following functions.

(a)

a

Example 1Find the limit of the f(x) as x approaches a for the following functions.

(a)

a

The limit does not exist as the function is not defined 'near' a.

Example 1 (cont'd)

(b)

a

Example 1 (cont'd)

(b)

a

The limit does not exist as the left side limit is not the same as right side limit.

Example 1 (cont'd)

(c)

f(a)

a

Example 1 (cont'd)

(c)

f(a)

a

The limit exists but it does not equal to f(a).

Example 1 (cont'd)

(d)

a

f(a)

Example 1 (cont'd)

(d)The limit exists and it equals to f(a).

a

f(a) )a(flimax

Evaluating LimitsIf f(x) is not broken at 'a', use direct substitution to evaluate its limit as x approaches 'a'

Otherwise, find the left side and right side limits and check if they are equal.

Example 2Evaluate the following limits if they exist.

(a) f(x) = 2x – 5 as x 1

f(x) is not broken at x = 1, so use direct sub.352)5x2(lim

1x

Example 2 (cont'd)(b) f(x) = ln x as x 0

f(0) is not defined. So consider limit from both sides.

But f(x) is not defined for x < 0.So the limit does not exist.

xlnlim0x

Example 2(c) f(x) = 1/(x – 2) as x 2

f(2) is not defined. So consider limit from both sides.

Since the left side limit does not equal to the right side limit, the limit of the f(x) as x approaches 2 does not exist.

)2x/(1limand)2x/(1lim2x2x

Example 2 (cont'd)(d) f(x) = (x – 1)/(x2 – 1) as x 1

f(x) = (x – 1)/(x + 1)(x – 1) = 1/(x + 1)1/(x + 1) is not broken at x = 1, so use direct sub.

5.01x

1xlim

21x

Example 2 (cont'd)

2)x(flimSo

2)7x(lim)x(flim

2)1x(lim)x(flim

3xif7x

3xif1x)x(f)e(

3x

2

3x3x

3x3x

2

Limits to InfinityIf f(x) = x + c, f(x) as x (note that x is the dominant term)

If f(x) = 1/x, f(x) 0 as x If f(x) = ax2 + bx + c, ax2 is the dominant term as x

)x

c

x

ba(xcbxax

222

Example 3Find the limit of f(x) as x (if they exist) for:

4x5x

2x7x3)x(f)d(

1xx

4x2x)x(f)c(

3x2x

x71)x(f)b(

1x4

3x2)x(f)a(

2

23

2

2

2

03x2x

x71limSo

xasx

7

x

x7

3x2x

x71)x(f)b(

2

1

1x4

3x2limSo

xas2

1

x4

x2

1x4

3x2)x(f)a(

2x

22

x

Example 3 (cont'd)

)x(flimand)x(flimSo

xasx3x

x3

4x5x

2x7x3)x(f)d(

11xx

4x2xlimSo

xas1x

x

1xx

4x2x)x(f)c(

xx

2

3

2

23

2

2

x

2

2

2

2

Example 3 (cont'd)

Trigonometric Limits

0xtanlim

0xsinlim

1xcoslim

0x

0x

0x

Example 4Find the following limits.

x

x4tanlim)d(

x

xcos1lim)c(

x

xtanlim)b(

x

xsinlim)a(

0x

0x

0x

0x

x

C

B

AO r

Consider the relationship between the areas OAC, sector OAC , and OAB

0

sinlimx

xInvestigating

x

x

C

B

AO r

Area of OAC = r2 sin x / 2Area of sector OAC = r2 x / 2Area of OAB = r2 tan x / 2

Example 4 (cont'd)

x

C

B

AO r

Area of OAC = r2 sin x / 2Area of sector OAC = r2 x / 2Area of OAB = r2 tan x / 2

So (size of areas)

r2 sin x / 2 < r2 x / 2 < r2 tan x / 2 sin x < x < tan x1 < x / sin x < 1 / cos x1 > sin x / x > cos x

Take limit as x 0 to get

That means sin x x as x 0

1x

xsinlim

0x

4x4

x4tanlim4

x4

x4tanlim4

x

x4tanlim)d(

0

01xcos1

xsin

x

xsinlim

)xcos1(x

xcos1lim

xcos1

xcos1

x

xcos1lim

x

xcos1lim)c(

1xcos

1

x

xsinlim

x

xtanlim)b(

0x40x0x

0x

2

0x0x

0x

0x0x

Example 4 (cont'd)

ContinuityGraphically a graph is continuous at x = a if it is not broken (disconnected) at that point.

Algebraically the limit of the function from both sides of 'a' must equal to f(a).

)a(f)x(flim)x(flimaxax

Example 5The following functions are not continuous at x = a.

a

af(a)

a

Why?

Example 6The following functions are continuous at x = a.

a

f(a)

a

Example 7Determine if the given function is continuous at the given point.

(a) f(x) = | x – 2 | at x = 2

(b) f(x) = x at x = 0

(c) f(x) = 1 / (x + 3) at x = -3

.3xatcontinuousnotis)3x/(1So3x

1limbut

3x

1lim)c(

.0xatcontinuousnotisxSo

0xfordefinednotisx)b(

.2xatcontinuousis|2x|So

)2(f0|2x|lim|2x|lim)a(

3x3x

2x2x

Example 7 (cont'd)

Example 8Given that f(x) is continuous over the set of all real numbers, find the values of a and b.

2xax2

2x16bx

1xax

)x(f

2

Only need to consider the junctions (x = -1 and x = 2)

1band4a

a46b2So

a4)x(flimand)x(flim6b2)2(f

6ba1So

6b)x(flimand)x(flima1)1(f

2x2x

1x1x

Example 8 (cont'd)

DifferentiabilityGraphical approach: A function f(x) is said to be differentiable at x = a if there is no 'corner' or 'vertical tangency' at that point.

A function must be continuous (but not sufficient) in order that it may be differentiable at that point.

Example 9The following functions are not differentiable at x = a.(a)

a

f(a)

Corner at x = a

Example 9 (cont'd)

(b)

y

x1 2 3 4 5 – 1 – 2 – 3 – 4

1

2

– 1

– 2

Vertical tangency at x = 1

Example 9 (cont'd)

(c)

y

x1 – 1 – 2 – 3 – 4 – 5

2

4

6

8

– 2

– 4

– 6

– 8

Not continuous (not even defined) at x = -2

Example 10The following functions are differentiable everywhere.

(a)

Example 10 (cont'd)

(b)

Example 10 (cont'd)

(c)

Derivative of a FunctionA function is differentiable at a point if it is continuous (not broken), smooth (no corner) and not vertical (no vertical tangency) at that point.

Its derivative is given by (First Principle)

h

xfhxfh

)()(lim

0

f(x)

f(x+h)

x x+h

P

Q

Differentiability (cont'd)

Differentiability (cont'd)The gradient of PQ is given by

As Q moves closer and closer to P (i.e. as h tends to 0), the limiting value of the gradient of PQ (i.e. the derivative of f(x) at x) becomes the tangent at P.

h

xfhxf )()(

Differentiability (cont'd)The derivative of a function y = f(x) is denoted by

It also represents the rate of change of y with respect to x.

dy dfor y or or f

dx dx

Example 11(a) Find the gradient function of y = 2x2 using first

principle. Find also the gradient at the point (3, 18).

(b)Use the definition (first principle) to find the derivative of ln x and hence find the derivative of ex.

1234dx

dy

x4

)h2x4(limh

h2hx4lim

h

x2)hhx2x(2lim

h

x2)hx(2lim

dx

dy)a(

3x

0h

2

0h

222

0h

22

0h

Example 11 (cont'd)

Example 11 (cont'd)

x

1

elnx

1

)h

xmwhere(])

m

11(lim[ln

x

1

)]x

h1(

h

xlim[ln

x

1)

x

h1(ln

h

1lim

hx

hxln

limh

xln)hx(lnlimxln

dx

d

m

m

0h0h

0h0h

Example 11 (cont'd)

xx

x

eedx

d

ydx

dy1

dx

dy

y

1

1ydx

dyln

dy

d1yln

dx

d

xylney

Example 12Find the derivative of the following functions from first principles.

(a) f(x) = 1/x(b) f(x) = x(c) f(x) = xn

Example 12 (cont'd)

2

0h

0h

0h

x

1

)hx(x

1lim

h)hx(xhxx

lim

hx1

hx1

lim)x

1(

dx

d)a(

Example 12 (cont'd)

x2

1xhx

1lim

)xhx(h

xhxlim

xhx

xhx

h

xhxlim

h

xhxlimx

dx

d)b(

0h

0h

0h

0h

Example 12 (cont'd)

1n

1n2n2n

1n

0h

n22n2n

1n

0h

nn

0h

n

nx

)h......hxCnx(lim

h

h......hxChnxlim

h

x)hx(limx

dx

d)c(

ConcavityIf f(x) opens downward, it is said to be concave down

If f(x) opens upwards, it is concave up

concave down concave up

f '(x): + 0 - - 0 +

Point of InflectionPoints of inflection: points where the curve changes from concave up to concave down or concave down to concave up

point of inflection

f '(x): - - - + + + maximum

Horizontal InflectionHorizontal inflection: a point of inflection where the graph is momentarily horizontal, dy/dx = 0

horizontal inflection

-ve

-ve+ve

+ve

Stationary PointsTurning points: max and min

Stationary points: max, min and horizontal inflection

dy/dx = 0

Example 13Given the f(x) graph below draw f '(x).

y

x

y

x

f '(x)f (x)

a b c

f '(x) changes as below:

x < a: +ve, but

x = a: f '(a) = 0 local max

a < x < b: -ve, , then (less –ve)point of inflection

Example 13 (cont'd)

y

x

f '(x)f (x)

a b c

x = b: f '(b) = 0 local min

b < x < c: +ve, , then point of inflection

x = c: f '(c) = 0 global max

x > c: -ve,

Example 13 (cont'd)

Noteworthy FeaturesMin TP: dy/dx = 0, sign change –ve, 0, +ve

Max TP: dy/dx = 0, sign change +ve, 0, –ve

Horizontal inflection: dy/dx = 0, +ve, 0, +ve or –ve, 0, –ve, (i.e. no sign change)

Point of inflection: d2y/dx2 = 0, (dy/dx is a max/min),

Example 14Graph the following function and its derivative.

Use your graphs to locate the stationary points and points of inflection on

y = x4/4 – 4x3/3 – 7x2/2 + 10x + 5and determine the nature of each.

y = x4/4 – 4x3/3 – 7x2/2 + 10x + 5dy/dx = x3 – 4x2 – 7x + 10

TP dy/dx = 0So x3 – 4x2 – 7x + 10 = 0i.e. (x + 2)(x – 1)(x – 5) = 0 x = -2, 1 or 5

When x = -2, y = -43/3When x = 1, y = 125/12When x = 5, y = -515/12

Example 14 (cont'd)

Example 14 (cont'd)

x-4 -2 2 4 6 8

y

-40

-30

-20

-10

10

20

f(x)

f '(x)

min

max

min PoI

PoI

Piecewise Defined FunctionsA piecewise defined function has different formulas for different parts of its domain.

At junction a filled circle indicates that a point actually exists there, whereas an empty circle shows a discontinuous point.

Example 15Given the function below

(a) Find f(-2), f(1) and f(2)

(b)Graph f and determine whether f is continuous at x = 0 and x = 2.

2xfor1x

2x0forx

0xfor1x/1

)x(f 2

2xfor1x

2x0forx

0xfor1x/1

)x(f 2

(a) f(-2) = 1/(-2) – 1 = -3/2f(1) = 12 = 1f(2) = 2 + 1 = 3

Example 15 (cont'd)

y

x1 2 3 4 – 1 – 2 – 3 – 4

1

2

3

4

5

6

– 1

– 2

– 3

– 4

Example 15 (cont'd)

2xatcontinuousnotis)x(f

)x(flim)x(flim

0xatcontinuousnotis)x(f

)x(flim)x(flim

2x2x

0x0x

Example 16Graph y = | x2 – 6x + 8 | and determine whether the function is continuous at x = 2 and x = 4.

Example 16 (cont'd)

x-1 1 2 3 4 5 6 7

y

-2

2

4

6

8

From graph, the function is continuous at x = 2 and x = 4.

The Sign FunctionIt can be considered as a logical function (especially in computer science)

It extracts the sign of the function

It returns 1 if f(x) is positive, 0 if f(x) equals to 0 and –1 if f(x) is negative.

0)x(f1

0)x(f0

0)x(f1

)x(sgn

Sign Function (cont'd)

0)x(f1

0)x(f0

0)x(f1

)x(sgn

x-3 -2 -1 1 2 3

y

-2

-1

1

2

Example 17Graph

(a) y = sgn (x/|x|)

(b)y = sgn (x2 – 1)

y

x1 2 3 – 1 – 2 – 3

1

2

– 1

– 2

(a) y = sgn (x/|x|)

Example 17 (cont'd)

Not continuous at x = 0

(b) y = sgn (x2 – 1)y

x1 2 – 1 – 2

1

2

– 1

– 2

Example 17 (cont'd)

Not continuous at x = -1 and x = 1

Greatest Integer FunctionAlso known as floor functionDefined as the greatest integer less than or equal to the numberThat is, it rounds any number down to the nearest integerSymbol: int [x] orint [4.2] = = 4 int [-2.1] = = -3

x

1.2 2.4

Greatest Integer Function (cont'd)y

x1 2 3 4 – 1 – 2 – 3

1

2

3

– 1

– 2

– 3

Not continuous at all integers.

Example 18Graph the following functions:

(a) int [2x – 1]

(b) int [x2]

(a) int [2x –1]

Consider 2x – 1 = n where n is an integerx = (n + 1) / 2So the 'breaking points' are steps of half of an integer

y

x1

2

1 3

2

2 5

2 –

1

2

– 1 –

3

2

– 2

1

2

3

– 1

– 2

– 3

– 4

Example 18 (cont'd)

(b) int [x2]

Consider x2 = n where n is a positive integerx = nSo the breaking points are square root of +ve integers

y

x1

2

1 3

2

2 5

2 –

1

2

– 1 –

3

2

– 2 –

5

2

1

2

3

4 Example 18 (cont'd)

Rules of Differentiation (Review)

2

1nn

gdxdg

fg)dxdf

()

g

f(

dx

d

dx

dgfg)

dx

df()gf(

dx

d

gdx

dfg

dg

d)]x(g[f

dx

d

gdx

df

dx

d)gf(

dx

d

xnxdx

d

Example 19Find the derivative of the following functions:

1xx3

)4x9(y)d(

)3x2(8x

5y)c(

)1x3x(

4y)b(

x

3x7y)a(

3

2

4

22

32

32'

22

22

)1x3x(

)3x2(8

)3x2()1x3x()2(4y

)1x3x(4y)b(

x

37x)1(37'y

x

3x7y)a(

Example 19 (cont'd)

2

3

3

3

2

4

3142

414

)8x(

)67x6()3x2(5

]8x

3x28[

8x

)3x2(5

8x

)3x2(40

)8x(

)3x2(5

)2()3x2(4)8x(5)3x2()8x)(1(5'y

)3x2()8x(5)3x2(8x

5y)c(

Example 19 (cont'd)

23

223

23

223

3

2

)1xx3(

)1x9()4x9()1xx3)(4x9(18

dx

dy

)1xx3(

)1x9()4x9()1xx3)(9)(4x9(2

dx

dy

1xx3

)4x9(y)d(

Example 19 (cont'd)

Differentiating ef(x) and ln f(x)

)x(f

)x(f)x(fln

dx

d

e)x(fedx

d

'

)x(f')x(f

Example 20Differentiate the following with respect to x:

x31x2

x

e2

1e2y)b(

ee3y)a(

x31x2

x31x2

x31x2x3

1x2

x

x

e2

3e4

dx

dy

e)2

3(e)2(2

dx

dy

e2

1e2

e2

1e2y)b(

e3dx

dy

ee3y)a(

Example 20 (cont'd)

Example 21Differentiate

x

2x

2

e1

exlney)b(

)3x7x(lne

1y)a(

]e1

e

ex

x2)

e1

ex(ln2[e

dx

dy

]e1

e

ex

x2[e)]e1(ln)ex([lne2

)]e1(ln)ex([lnee1

exlney)b(

)3x7x

7x2(

e

1

dx

dy

)3x7x(lne

1y)a(

x

x

2x

2x2

x

x

2x2x2x2

x2x2x

2x2

2

2

Example 21 (cont'd)

Differentiability (Revisit)Graphical approach: continuous, no corner, no vertical tangency

Algebraical approach:

)gencytanverticalno(finiteis)a(f

)cornerno()a(f)x(flim)x(flim

)continuous()a(f)x(flim)x(flim

'

''

ax

'

ax

axax

Example 22Determine if the following functions are differentiable at the indicated points.

(a) y = 1 / (x + 1) at x = -1

(b) y = | x + 1 | at x = -1

(c) f(x) = -6x + 5 for x < 3 = -x2 – 4 for x 3 at x = 3

(a) Let f(x) = 1 / (x + 1)

f(-1) is not definedf(x) is not continuous at x = -1f(x) is not differentiable at x = -1

Example 22 (cont'd)

1xatabledifferentinotis)x(f

1)x('flim1)x('flim

1x1

1x1)x('f

1xatcontinuousis)x(f

0)1(f)x(flim)x(flim

1x1x

1x1x)x(f

|1x|)x(fLet)b(

1x1x

1x1x

Example 22 (cont'd)

3xatabledifferentiis)x(f

6)x('flim)x('flim

3xx2

3x6)x('f

3xatcontinuousis)x(f

13)3(f)x(flim)x(flim

3x4x

3x5x6)x(f)c(

3x3x

3x3x

2

Example 22 (cont'd)

Example 23Find the value of a and b so that

f(x) = 3x + 1 for x < 1 = x2 + ax + b for x 1

is continuous and differentiable everywhere.

Possible discontinuity and non-differentiability at x = 1

continuous at x = 1 if 4 = 1 + a + bi.e. a + b = 3

f ’(x) = 3 for x < 1 = 2x + a for x > 1differentiable at x = 1 if 3 = 2 + a

So a = 1 and b = 2

Example 23 (cont'd)

Riemann SumsTo find an approximate area under a curve between two x values [a, b]

The area is divided into n rectangles of equal width

So the width x = (b – a)/n

There are many ways to find the height h of each rectangle (see later)

Then the required area A = hx over the interval [a, b]

Example 24The shaded area below shows the exact area under the curve f(x) = x3 – 3x2 + 8 in the interval [0, 3]

x0.5 1 1.5 2 2.5 3

y

1

2

3

4

5

6

7

8

9

y = x3 – 3x2 + 8

Actual area = 17.25

Example 25(Example 24) Consider n = 5, x = (3 – 0)/5 = 0.6 and h = left endpoint

x = (3 – 0)/5 = 0.6

Left endpoint for the 3rd rectangle

Example 25 (cont'd)f(x1)= (0)3 – 3(0)2 + 8 = 8f(x2)= (0.6)3 – 3(0.6)2 + 8 = 7.136f(x3)= (1.2)3 – 3(1.2)2 + 8 = 5.408f(x4)= (1.8)3 – 3(1.8)2 + 8 = 4.112f(x5)= (2.4)3 – 3(2.4)2 + 8 = 4.544

52.17

)544.4112.4408.5136.78(6.0

x)x(fA5

1ii

Example 26(Example 24) Consider n = 5, x = (3 – 0)/5 = 0.6 and h = right endpoint

Right endpoint for the 3rd rectangle

Example 26 (cont'd)f(x1)= (0.6)3 – 3(0.6)2 + 8 = 7.136f(x2)= (1.2)3 – 3(1.2)2 + 8 = 5.408f(x3)= (1.8)3 – 3(1.8)2 + 8 = 4.112f(x4)= (2.4)3 – 3(2.4)2 + 8 = 4.544f(x5)= (3.0)3 – 3(3.0)2 + 8 = 8

52.17

)8544.4112.4408.5136.7(6.0

x)x(fA5

1ii

Example 27(Example 24) Consider n = 5, x = (3 – 0)/5 = 0.6 and h = minimum point

Minimum point for the 4th rectangle

Example 27 (cont'd)f(x1)= (0.6)3 – 3(0.6)2 + 8 = 7.136f(x2)= (1.2)3 – 3(1.2)2 + 8 = 5.408f(x3)= (1.8)3 – 3(1.8)2 + 8 = 4.112 f(x4)= (2.0)3 – 3(2.0)2 + 8 = 4 f(x5)= (2.4)3 – 3(2.4)2 + 8 = 4.544

12.15

)544.44112.4408.5136.7(6.0

x)x(fA5

1ii

Example 28(Example 24) Consider n = 5, x = (3 – 0)/5 = 0.6 and h = maximum point

Maximum point for the 4th rectangle

Example 28 (cont'd)f(x1)= (0)3 – 3(0)2 + 8 = 8 f(x2)= (0.6)3 – 3(0.6)2 + 8 = 7.136f(x3)= (1.2)3 – 3(1.2)2 + 8 = 5.408f(x4)= (2.4)3 – 3(2.4)2 + 8 = 4.544f(x5)= (3.0)3 – 3(3.0)2 + 8 = 8

8528.19

)8544.4408.5136.78(6.0

x)x(fA5

1ii

Example 29(Example 24) Consider n = 5, x = (3 – 0)/5 = 0.6 and h = midpoint

Midpoint for the 3rd rectangle

Example 29 (cont'd)f(x1)= (0.3)3 – 3(0.3)2 + 8 = 7.757f(x2)= (0.9)3 – 3(0.9)2 + 8 = 6.299f(x3)= (1.5)3 – 3(1.5)2 + 8 = 4.625f(x4)= (2.1)3 – 3(2.1)2 + 8 = 4.031f(x5)= (2.7)3 – 3(2.7)2 + 8 = 5.813

115.17

)813.5031.4625.4299.6757.7(6.0

x)x(fA5

1ii

A Better ApproximationDue to the use of h (left, right, mid, min and max), the rectangles do not truly represent the area under the curve for each strip

If n (number of rectangles) increases, the error decreases

Example 30n = 20, x = (3 – 0)/20 = 0.15, h = midpoint

Actual area = 17.25

Example 31n = 100, x = (3 – 0)/100 = 0.03, h = midpoint

Actual area = 17.25

Limit of a SumThe more rectangles, the greater accuracy

So the actual area A is given by

This is written as

x)x(flimAn

egrationintcalledisprocessabovetheand

sumaofitlimtherepresentswhere

dx)x(fx)x(flimAn

Integration f(x)dx represents the area under the curve

x1 2 3 4

y

1

2

3

4

5

6

7

8

4

1dx)x(f

f(x)

Example 32

b 3 4 5 6 7 x

Area

Evaluate the area under the curve y = 3 from x = 2 to x = b by completing the following table.

Hence give an answer for where k is a constant

x

akdx

Example 32 (cont'd)

b 3 4 5 6 7 b

Area 3 6 9 12 15 3(b – 2)

x1 2 3 4 5 6 7 8 9 10

y

1

2

3

4

)ab(kkdxb

a

Evaluate the area under the curve y = 2x from x = 0 to x = b by completing the following table.

Hence give an answer for where k is a constant

Example 33

b 0 1 2 3 4 b

Area

b

0kxdx

Example 33 (cont'd)

b 0 1 2 3 4 x

Area 0 1 4 9 16b2b/2 =

b2

x1 2 3 4 5 6

y

2

4

6

8

10

12

14

2

kbkxdx

2b

0

Given the areas under the curve y = x2 from x = 0 to x = 4 in the following table, find the area when x = b

Hence give an answer for

Example 34

b 0 1 2 3 4 b

Area 0 1/3 8/3 9 64/3

b

0

2dxx

Example 34 (cont'd)

b 0 1 2 3 4 b

Area 0 1/3 8/3 9 64/3 b3/3

x1 2 3 4 5

y

5

10

15

20

25

30

3

bdxx

3b

0

2