More Review for Test I

Calculation of Limits

limx 4

x 2x 4

limx

x3

x

4 x3

x2

2

f {x

2 x 3

7 x otherwise

limx 2

f(x) limx 3

f(x)lim +x 3

f(x)lim -x 3

f(x)

Continuity

Analytic Definition: the function f is continuous at x = a if

When you can calculate limits simply by ‘plugging in’ .

limx a

f(x) f(a)

Geometric Definition: the function f is continuous at x = a if the graph of f is “not broken” at a.

lim +x a

f(x) lim -x a

f(x) = f(a)

f {x

2 x 3

7 x otherwise

limx 3

x2

9

f(x) x2

is continuous at x = 3

= f(3)

is not continuous at x = 3

limx 3

f(x)does not exist.

f {x

2 x 3

W x otherwiseIs continuous at 3 if W is chosen right

lim -x 3

f(x)= 9

lim +x 3

f(x) = W + 3

They are equal exactly when W = 6

f(3) = W + 3

Definition of the Derivative

Geometric Definition:

If f(x) is a function then f ‘ (a), the derivative of f at x= a is the slope of the tangent line to the graph of f(x) at (a, f(a))

Analytic or “limit” Definition

f '(a) limh 0

f(a+h) - f(a)

h

Derivative Rules

For exam are responsible for:

1. (cf) ‘ = c f ‘2. (f + g) ‘ = f ‘ + g ‘3. (fg) ‘ = f ’g + f g ‘ 4. ( ) ‘ =

5. ( ) ‘ =

xn

n x( )n 1

f

g

f ' g f g '

g2

Calculations with tangent lines

1. Tangent line to graph of f(x) at x = a

y y0 m ( )x x0 General Eqn For Line

y ( )f a m ( )x a m = f ‘(a)

y f(a) f '(a) ( )x a

Eqn of tangent to graph of at x = 2

a = 2, f(2) = -2, f ' (x) 3 x2

4

( )f x x3

5 x

f '(x) 3 x2

5 f ‘(2) = 7

Eqn: y +2 = 7(x-2)

Use of Rules:

Suppose f(7) = 3 and f ‘(7) = 4. What is the equation of the tangent line to the graph of f(x) at x = 7?

y – f(7) = f ‘(7) (x-7)

y – 3 = 4 (x-7)

What is the equation of the line through (7,3) which is perpendicular to the tangent line to the graph of f(x) at x = 7?

y – 3 = -1/4 (x-7)

Suppose f(9) =2, f ‘(9) = 4, g (9) = 8, g ‘(9) = 7.What is the equation of the tangent line to the graph ofh(x) = f(x) – g(x) at x = 9?

y – h(9) = h ‘(9) (x-9)

h(9) = f(9) – g(9) = 2-8 = -6

h ‘(9) = f ‘(9) – g ‘(9) = 4 – 7 = -3

y +6 = (-3)(x-9)

E q n o f T a n g e n t L i n e a t x = 3

f ‘ ( a ) = f ‘ ( 3 ) i s t h e s l o p e o f t h i s t a n g e n t l i n e o r a p p r o x i m a t e l y- 3 / 2 ( f r o m g r a p h )

y a f ' (a) ( )x ( )f a

H e r e :a = 3 , f ( a ) = - 3 ( f r o m g r a p h )

P o i n t ( a , f ( a ) ) = ( 3 , - 3 )

E q u a t i o n o f t a n g e n t l i n e i s y – ( - 3 ) ) = ( - 3 / 2 ) ( x - 3 )

M a y b e p u t i n s t a n d a r d f o r m s u c h a s y = 3 / 2 – ( 3 / 2 ) x .

Tangent Data From Graph

What is the equation of the tangent line to the graph of h(x) = xf(x) at x = 3? y + 3 = h ‘(3)(x-3), h’(x) = f(x) + x f ‘(x). h ‘(3) = …

Interpretation of the Tangent Line

Philosophically, the tangent line to the graph of f(x) at (a, f(a)is the line that best approximates the graph of f “near(a,f(a))”

Given f(x), and line y = mx + bhow can you tell if the lineis the tangent line at x = a?

Answer:(1) y = f(a) when x = a ,(2) y ‘ (a) = f ‘(a) (since y ‘ (a) is the slope of the line and the slope of the line is f ‘(a))

Question: Is y = 9x -19 the tangent line to the graph of

f(x) = at x = 4?

Answer=NO When x = 4, y = 9*4-19 = 17 = f(4), but y ‘ (4) = 9 while f ‘(4) = 7

x2

x 5

Question: Suppose f ‘ (3) = 2 and f(3) = 5, g (3) =7 and g ‘(3) = 4. What is the equation of the tangent line to the graph of

at x = 3?

( )h xf(x)

g(x)

SLN: To know the tangent line at x =3 we need to know h(3) and h ‘(3)

( )h xf(3)

g(3)

( )h ' xf '(x)*g(x) - f(x)*g '(x)

g(x)2

( )h ' 32*7 - 5*4

72

-6

49=

( )h 35

7

y – h(3) = h ‘(3)*(x-3) is the equation

The equation of the tangent line to the graph of f(x) at x = 5 is

y = 14*x-32. What are f(5) and f ‘(5).Ans: The point (5, f(5)) is on the graph of f and on the tangent line.When x = 5, y = 14*5-32 = 38 so this has to be f(5).The slope of the line is y ‘= 14 so this has to be f ‘(5).

The equation of the tangent line to the graph of h(x) = x*f(x)at x = 7 is y=5*x -3. What are f ‘(7) and f(7)?ANS: h ‘(x) = f(x) + x*f ‘(x) and y ‘(7) = 5 so 5 = h ‘(7) = f(7) + 7*f ‘(7).When x=7 y = 5*7-3=32 so 32 = h(7) = 7*f(7)

Thus we have equations: 5 = f(7) + 7* f ‘(7) 32 = 7*f(7) which are easily solved for f(7) and f ‘(7).

The derivative and Piecewise Functions

If f is defined piecewise then f ‘(x) for x other than end points is calculated piecewise itself. Generally if f is continuous at an end pointand the derivatives exist on the pieces and the limits of the derivatives exist (and are equal) from the left and the right then this is the derivative at the end points.

f(x) {x

21 x 2

7 x otherwise

If

Except at x = 2 we calculate piecewise.

f '(x) {2 x x 2

-1 2 x

f ‘(2) DNE since limit from the left is 4but limit from the right is -1

Suppose

Can C be chosen so that f ‘(x) has a derivative at x = 2?If f ‘(2) exists then f has to be continuous at x = 2. This meansthe limits from the right and left have to exist and be equal to f(2).

( )f x

x2

x 1 x 2

C x3 2 x

= 3,

so C must = 3/8 if f is to be continuous.

( )f x

x2

x 1 x 2

5 x3

82 x

= C8lim -x 2

( )f x lim -x 2

x2

x 1lim

+x 2( )f x lim

+x 2C x

3

f '(x)

2 x 1 x 2

15 x2

82 x

From the left f ‘=3, from the right f ‘= 60/8 – f ‘DNE