lesson 2.1 the derivative and the tangent line problem quiz

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What does it mean to say that a line is tangent to a curve at a point? . P For a circle, the tangent line at a point P is the line that is perpendicular to the radial line at point P. For a general curve, however, the problem is more difficult.

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Lesson 2.1The Derivative and the Tangent Line Problem

Quiz

What does it mean to say that a line is tangent to a curve at a point?

.. P For a circle, the tangent line at a point P is the line that is perpendicular to the radial line at point P.

For a general curve, however, the problem is more difficult.

Essentially, the problem of finding the tangent line at a point P boils down to the problem of finding the slope of the tangent line at point P. You can approximate this slope using a secant line through two points on the curve.

(c+x, f(c+ x).

.(c, f(c))-------------

---------------x

y

0 0 0

Find the slope of 2 3 at 2,1 .

2 2 2 2 3 1 2lim l i 2im l mx x x

f x x

f x f x xx x x

2.1 The Derivative and the Tangent Line Problem

0

limx

f c x f cm

x

2.1 The Derivative and the Tangent Line Problem

0

limx

f c x f cm

x

2

2 2 2 2 2

0 0

2

0 0

Find the derivative of +1.

1 1 2lim lim

2lim lim 2 2

x x

x x

f x x

x x x x x x x xx x

x x x x x xx

0

' limx

f x x f xf x

x

The slope of a function is its

derivative.

2.1 The Derivative and the Tangent Line Problem

2

2 2

0

Find the slope of +1 at 0,1 & 1,2 .

1 1lim

' 0 2 0

' 1

2

0

2 1 2

x

f x x

x xx

x

xf

f

2.1 The Derivative and the Tangent Line Problem

'f x

dydx

'y d f x

dx

xD y

2.1 The Derivative and the Tangent Line Problem

3 2 ' ?f x x x f x

0' lim

x

f x x f xf x

x

3 3 2 2 3

0 0

3 2 2 2 2 3 3

0

2 2 32 2 2

0 0

2 2 2 2 2 2lim lim

2 2 2 2 2lim

3 3 2lim lim 3 3 2 3 2

x x

x

x x

x x x x x x x x x x x x x x x x

x xx x x x x x x x x x x x x x

xx x x x x x x x x x x

x

2' 3 2f x x

3 2 22a b a b a ab b

Complete onWhiteboard

2.1 The Derivative and the Tangent Line Problem

' ?f x x f x

0' lim

x

f x x f xf x

x

0limx

x x xx

1'2

f xx

0

0

0

lim

lim

1lim 12

x

x

x

x x x x x x

x

x x x xx x x

x x x x

x x x

Complete onWhiteboard

2.1 The Derivative and the Tangent Line Problem

Find the slope of at 1,1 and 4,2 . Discuss the

behavior of at the origin.

f x x

f x

1'2

f xx

1' 1

2121

f

1' 42

144

f 0

1 1lim02x x

has a vertical tangent @ 0,0 .f0x

2.1 The Derivative and Tangent Line Problem

is differentiable on a,b

is a continuous, smooth curve on , and

does not have a vertical tangent.

f

fa b

f

AP EXAM

2If , find '.y yt

0

0

0

0

20

20 2

' lim

2 / 2 /lim

22

lim

2 21lim

1 2lim

2li 2m

t

t

t

t

t

t

f t t f ty

tt t t

tt tt

t t t t t tt

t t tt t

t

t t

tt t t t

t t t

Differentiability and continuityThe following alternative limit form of the derivative

is useful in investigating the relationship between differentiability and continuity. The derivative of f at c is

' lim

x c

f x f cf c

x c

}

2.1 The Derivative and the Tangent Line Problem

2 2

2 2

2 02lim lim 1

2 22 02

lim lim 12 2

x x

x x

xf x fx x

xf x fx x

' 2f DNE

2 Find ' at (2,0).f x x f x ' lim

x c

f x f cf c

x c

2.1 The Derivative and the Tangent Line Problem

1/ 3 Find ' at 0.f x x f x x

0

1/3

0

2/30

0lim

00lim

01lim tangent is vertic l ' 0a

x

x

x

f x fx

x

xNE

x

f D

Vertical Tangent LineIf a function is continuous at a point c and , then x = c

is a vertical tangent line for the function.

limx c

f x f cx c

2.1 The Derivative and the Tangent Line Problem

THM 2.1Differentiability Continuity

HW 2.1/3,4,5-15odd,16,21,24,25,27-32,33,35,37,41,45,47,62

x 0 x 0

x

2

0 x 0

x 0

113. ' ?1

1 11 1' lim lim

1 11 1lim lim1 1 1 1

1lim1 1

11

f x f xx

f x x f x x x xf xx x

x x x xx x x x x x x x

x x x x

x 0 x 0

x 0

x 0

x 0

x 0

116. ' ?

1 1

' lim li

1

m

1lim

1lim

1lim

1l m2 2

1i

f x f xx

f x x f x x x xf xx x

x x xx x x x

x x x x x xx x x x x x x

x x xx x x x x x x

x x x x x x xx x x

x 0 x 0

x 0 0

0 0

(

121. 1, 2

Find the tangent to @ .1 1

' lim lim

1 1 1 1lim lim

1lim l

)

im

1,2

x

x x

f x xxf

x x xf x x f x x x xf xx x

x x x x x x xx

x x x x x x x x

x x x x x x x xx x x x

1 1

2

2

1

0

2 0 1 2

1

0' 11

x x x

f

y

x

x

y

y m x

x y

x

3 / 2

3 / 23 / 2 3 / 2

124.

Find the tangent to that is to 2 6 0.1' (Earlier Problem)

21 1 1 1 1 1

2211 1 We need the line through 1,1 with 2

11 12

f xx

f x y

f xx

x xx

y x

x

f m

2 2

2 21 1 1 1

21

1 1 11 0 1lim lim lim lim1 1 1 1 1 1

1 2lim vertical tangent01

The limit from the right DNE

since is undefined for 1.is not different ab i le

x x x x

x

f x f x xx xx x x x x x

x

x

ff x

at 1.x

2

Find the derivative from the left and the right at 1.

Is the function differentiable at 11?

x

x f x x

2 Use the definition of derivative to find ' if 2 1.f x f x x x

0

' limx

f x x f xf x

x

2 2

0 0

2 2 2

0

2 2 2

0

2

0 0

2 1 2 1' lim lim

2 2 1 2 1lim

2 4 2 1 2 1lim

4 2lim lim 1 14 2 4

x x

x

x

x x

x x x x x xf x x f xf x

x xx x x x x x x x

xx x x x x x x x

xx x x x x x

xx

Yea! You finished the lesson!

Now get to work!

(c+x, f(c+ x)..(c, f(c)) ----------

------------

x

y

.(c+x, f(c+ x).

(c, f(c))-------------

---------------

x

y

.(c, f(c))

If f is defined on an open interval containing c, and if the limit

exists, then the line passing through (c, f(c)) with slope m is the tangent line to the graph of f at point (c, f(c)).

lim

x→0

f(c + x) – f(c)

x= m

:4 1 4 3

4. (a) 4 1 4 3

4 1 (b) ' 1

4 1116.

2x x1 324. 2 2

28. (e)30. (a)32. (d)62. '

Even Answersf f f f

f ff

y x

f x DNE

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