Transcript Aim: What do slope, tangent and the derivative have to do

Aim: What do slope, tangent and the
derivative have to do with each other?
Do Now: What is the equation of the line
tangent to the circle at point (7, 8)?
10
8
6
4
2
5
10
-2
Aim: The Tangent Problem & the Derivative
Course: Calculus
Tangents & Secants
A tangent to a circle is a line in the plane of
the circle that intersects the circle in exactly
one point.
A
B
O
C
A secant of a circle is a line that intersects the
circle in two points.
Aim: The Tangent Problem & the Derivative
Course: Calculus
radius = 1
center at (0,0)
Tan 
1 y
cos , sin 
(x,y)
1
-1

cos 
1
length of the leg opposite
tan 
length of the leg adjacentto 
-1
sin
y
tan 

 slope
cos x
Aim: The Tangent Problem & the Derivative
Course: Calculus
x
Tangents to a Graph
(x3, y3)
(x2, y2)
(x4, y4)
(x1, y1)
slope is
steep!
Unlike a tangent to a circle,
tangent lines of curves can
intersect the graph at more than
one point.
Aim: The Tangent Problem & the Derivative
slope is
level:
m=0
slope is
falling:
m is (-)
3
2
1
2
A
-1
Course: Calculus
Finding the Slope (tangent) of a Graph at a Point
h x  = x 2
10
8
6
4
2
2
(1, 1)
1
5
y 2
slope m 
 2
x 1
This is an approximation. How can we be
sure this line is really tangent to f(x) at (1, 1)?
Aim: The Tangent Problem & the Derivative
Course: Calculus
Slope and the Limit Process
3
2.5
(x + h, f(x + h))
2
A more precise method
for finding the slope of
the tangent through (x,
f(x)) employs use of
the secant line.
f(x + h) – f(x)
1.5
1
slope msec
x, f(x)
0.5
h
1
h is the change in x
f(x + h) – f(x) is the
change in y
slope msec
y

x
f ( x  h)  f ( x )

h
This is a very rough
approximation of the
slope of the tangent at
the point (x, f(x)).
Aim: The Tangent Problem & the Derivative
Course: Calculus
Slope and the Limit Process
3
slope msec
2.5
2
1.5
(x + h, f(x + h))
1
y

x
h is the change in x
f(x + h) – f(x) is the
change in y
f(x + h) – f(x)
x, f(x)
0.5
h
1
As (x + h, f(x + h))
moves down the curve
and gets closer to
(x, f(x)), the slope of
the secant more
approximates the
slope of the tangent at
(x, f(x).
Aim: The Tangent Problem & the Derivative
Course: Calculus
Slope and the Limit Process
3
slope msec
2.5
y

x
h is the change in x
f(x + h) – f(x) is the
change in y
2
1.5
(x + h, f(x + h))
f(x + h) – f(x)
1
x, f(x)
0.5
h
1
What is happening to
h, the change in x?
It’s approaching 0,
or its limit at x as h
approaches 0.
Aim: The Tangent Problem & the Derivative
Course: Calculus
Slope and the Limit Process
3
2.5
2
1.5
1
0.5
1
As h  0, the slope of the
secant, which
approximates the slope
of the tangent at (x, f(x))
more closely as (x + h, f(x
+ h)) moved down the
curve. At reaching its
limit, the slope of the
secant equaled the slope
of the tangent at (x, f(x)).
slope m tan  lim msec
h 0
slope msec 
f ( x  h)  f ( x )
h
Aim: The Tangent Problem & the Derivative
Course: Calculus
Definition of slope of a Graph
The slope m of the graph
of f at the point (x, f(x)) ,
is equal to the slope of its
tangent line at (x, f(x)),
and is given by
3
2.5
2
1.5
slope m tan  lim msec
1
h 0
provided this limit exists.
0.5
1
f ( x  h)  f ( x )
m  lim
h 0
h
difference quotient
Aim: The Tangent Problem & the Derivative
Course: Calculus
Model Problem
Find the slope of the graph f(x) = x2 at the
point (-2, 4).
f ( x  h)  f ( x )
m  lim
h 0
f ( 2  h)  f ( 2)
h 0
h
( 2  h)2  ( 2)2
 lim
h 0
h
4  4h  h 2  4
 lim
h 0
h
 4h  h 2
 lim
h 0
h
h( 4  h)
 lim
h 0
h
m  lim
m
m
m
m
h
set up difference
quotient
q x  = x 2
Use f(x) = x2
8
Expand
6
Simplify
4
D: (-2.00, 4.00)
D
Factor and divide out
2
m  lim( 4  h)
Simplify
m  lim( 4  h)  4
Slope ED = -4.00
h 0
h 0
E
Evaluate the limit
Aim: The Tangent Problem & the Derivative
Course: Calculus
Slope at Specific Point vs. Formula
What is the difference between the
following two versions of the difference
quotient?
f ( x  h)  f ( x )
(1) m  lim
h 0
h
f ( c  h)  f ( c )
( 2) m  lim
h 0
h
(1) Produces a formula for finding the
slope of any point on the function.
(2) Finds the slope of the graph for the
specific coordinate (c, f(c)).
Aim: The Tangent Problem & the Derivative
Course: Calculus
Definition of the Derivative
The function found by evaluating the limit of
the difference quotient is called the
derivative of f at x. It is denoted by f ’(x),
which is read “f prime of x”.
The derivative of f at x is
f ( x  h)  f ( x )
f ' ( x )  lim
h 0
h
provided this limit exists.
The derivative f’(x) is a formula for the
slope of the tangent line to the graph of
f at the point (x,f(x)).
Aim: The Tangent Problem & the Derivative
Course: Calculus
Finding a Derivative
Find the derivative of f(x) = 3x2 – 2x.
f ( x  h)  f ( x )
f '( x )  lim
h 0
h
[3( x  h)2  2( x  h)]  ( 3 x 2  2 x )
f ' ( x )  lim
h 0
h
3 x 2  6 xh  3h2  2 x  2h  3 x 2  2 x
f ' ( x )  lim
h 0
h
6 xh  3h2  2h
f ' ( x )  lim
h 0
h
h(6 x  3h  2)
f ' ( x )  lim
factor out h
h 0
h
f ' ( x )  lim(6 x  3h  2)  6 x  2
h 0
Aim: The Tangent Problem & the Derivative
Course: Calculus
Aim: What is the connection between
differentiability and continuity?
Do Now:
Find the equation of the line tangent to
f ( x)  2 x
Find the slope of tangent at x = 9
Aim: The Tangent Problem & the Derivative
Course: Calculus
Differentiability and Continuity
What is the relationship, if any, between
differentiability and continuity?
f(x) is a continuous function
7
(x, f(x))
6
f(x) – f(c)
(c, f(c))
5
4
x–c
Is there a limit as x
approaches c? YES
3
2
1
x
c
f ( x )  f (c) alternative
form of
f '(c)  lim
x c
derivative
xc
0.5
1
1.5
Aim: The Tangent Problem & the Derivative
2
2.5
Course: Calculus
3
Differentiability and Continuity
3.5
Is this step function
differentiable at x = 1?
3
2.5
By definition the derivative is
a limit. If there is no limit at x
= c, then the function is not
differentiable at x = c.
2
1.5
1
0.5
-1
1
-0.5
2
3
4
5
f ( x)  [[ x]]
-1
Does this step function, the greatest
integer function, have a limit at 1?
NO: f(x) approaches a different number from
the right side of 1 than it does from the left side.
Aim: The Tangent Problem & the Derivative
Course: Calculus
Differentiability and Continuity
If f is differentiable at x = c, then f is
continuous at x = c.
Is the Converse true?
NO
If f is continuous at x = c, then
f is differentiable at x = c.
Aim: The Tangent Problem & the Derivative
Course: Calculus
3.5
Graphs with Sharp Turns – Differentiable?
3
f ( x )  f (c) alternative
form of
f '(c)  lim
x c
derivative
xc
m = -1 f(x) = |x – 2| m = 1
2.5
2
1.5
1
0.5
-1
1
2
3
4
5
6
-0.5
Is this function continuous at 2? YES
Is this function differentiable at 2?
-1
-1.5
lim | x  2 | ? lim f ( x)  f (2)  lim x  2  0  1
x2
x  2
x  2
x2
x2
x2 0
lim | x  2 | ? lim f ( x)  f (2)  lim
 1


x2
x 2
x 2
x2
x2
One-sided limits are not equal, f is therefore not
differentiable
2. There
tangent
line at (2, 0)
Aim: at
The Tangent
Problem &is
theno
Derivative
Course: Calculus
Graph with a Vertical Tangent Line
1.2
f(x) = x1/3
1
0.8
Is f continuous at 0?
0.6
0.4
YES
0.2
Does a limit exist at 0?
-1
1
-0.2
1
lim x 3 ?
-0.4
x 0
-0.6
f ( x)  f (0)
lim
x 0
x0
lim
x 0
x
1
0
x
3
-0.8
-1
lim
x 0
1
x
2
3
 UND
NO
f is not differentiable at 0; slope
of vertical line is undefined.
Aim: The Tangent Problem & the Derivative
Course: Calculus
Differentiability Implies Continuity
corner
vertical
tangent
a
f is not
continuous at a
therefore not
differentiable
b
c
f is
continuous at
b & c, but not
differentiable
Aim: The Tangent Problem & the Derivative
d
f is continuous at
d and
differentiable
Course: Calculus
Summary
Aim: The Tangent Problem & the Derivative
Course: Calculus