W-upGet out note paperFind 12.3 notesWrite the 3 steps to determine if a

function is continuous

1) F is defined at c; that is, c is in the domain of f so that f(c) equals a #.

2)

3) *So the same output is approached from both sides.

14.1 Horizontal and Vertical Tangent Lines; Continuity and Differentiability• Find horizontal and vertical tangent lines• Discuss the graph of a function f where the

derivative of f does not exist

Horizontal tangent lineIf function is differentiable at point c,

horizontal tangent line occurs when f ‘ (c) = 0

Ex1: At what points is the tangent line of f(x)=x3 + 3x2 – 24x horizontal?

Find derivative: f ‘(x) = 3x2 + 6x – 24Set f ‘(x) = 0 and solve 0 = 3x2 + 6x – 24 factor

3(x2 + 2x – 8) = 0 (x + 4 ) (x – 2) = 0 x = -4 and x = 2 Horizontal tangent lines will occur at (-4,80) and (2, -

28)

Plug x values into ORIGINAL equation to find the y-value where tangent line is horizontal• f(-4) = (-4)3 + 3(-4)2 – 24(-4)

= 80• f(2) = (2)3 + 3(2)2 – 24(2) = -

28

Conditions for Vertical Tangent LinesIf a vertical tangent line is present at a point

(c, f(c)) on a continuous function then1) x = c is in the domain2) f’(x) must be unbounded at x = c

Unbounded will be where the denominator of the derivative = 0

Step 1 state domain – all real numbers

Step 2- find derivative f ‘(x) = -6x + 12To find horizontal set f ‘(x) = 0 and

solve0 = -6x + 12x = 2Plug 2 into original to find y f(2) = -3(2)2 + 12(2) = 12Horizontal tangent line at (2, 12)

Ex1: find horizontal and vertical tangent lines of f(x) = -3x2 + 12x

When will there be a vertical tangent line?

NO VERTICAL TAN LINE – there is no denominator (no unbounded points)

Ex2: find horizontal and vertical tangent lines of

1) FIND domain 2) find the derivative – good times

2/3

( )1

xf x

x

2/31/3

2

2 1

31

xx

xx

1/3 2/3

2

213'( )1

x x xf x

x

Get common denominator

1/3

2

2 1 3

31

x x

xx

1

21

1x

21/3

2'( )

3 1

xf x

x x

Find horizontal tangent line set derivative = 0 and solve

A fraction will be zero when numerator is zero

x=-2Find y by plugging x into

original equation

Horizontal tangent line at (-2,.53)

To find vertical tangent line set denominator of derivative = 0 and solve

Set each factor = 0 and solve

x = 0 and x = 1Throw out 1 NOT IN

DOMAINFind y by plugging x into

original equation

Vertical tangent line a (0,0)

21/3

2'( )

3 1

xf x

x x

21/3

203 1

x

x x

2/3

( )1

xf x

x

2/32( 2)

2 10.53f

21/30 3 1x x

21/30 3 1 0x and x

2/30(0

10)

0f

Continuity and DifferentiabilityIf c is a number in the domain of f and f is

differentiable at the number c, then f is continuous at cIf it is not continuous it is not differentiable.

Converse is not always true – if a graph is continuous at c, it may not be differentiable

Given: a) determine if it is continuous at x = 0Is it defined find left and right side limits

Yes continuous all equal zeroB) does f’(0) exist? Plug in zero

2/5( )f x x

2/5(0) 0 0f

2/5 2/5

0 0lim 0 lim 0x x

x x

33/5

/52'( )

5

2

5f x x

x

3/5

2'(0) .

5

2. .

0(0)Ef D N Will this be a VERTICAL

tangent line or NO tangent line?

Given: a) determine if it is continuous at x = 1

2

6 1( ) 1

5 1

x if xf x at x

x if x

Is it defined f(1) = 12 + 5 = 6 find left and right side limits

Yes continuous all equal six

2

1 1lim 1 5 6 lim 6(1) 6x x

B) does f’(c) exist? GRAPH2

6 1( ) 1

5 1

x if xf x at x

x if x

1 6

0 0

-1 -6

1 6

2 9

3 14

f(x) = x2+5; x> 1

f(x) = 6x; x< 1

• f ‘(1) does not exist, there are two different slopes at 1

• Find the derivative of each part• f’ (x) = 6 f’(x) = 2x

f’(1) = 6 f’ (1) = 2

Homework14.1 # 1 – 33 oddsDon’t forget to graph # 27-33